There is a road with length l meters. The start of the road has coordinate 0, the end of the road has coordinate l.

There are two cars, the first standing at the start of the road and the second standing at the end of the road. They will start driving simultaneously. The first car will drive from the start to the end and the second car will drive from the end to the start.

Initially, they will drive with a speed of 1 meter per second. There are n flags at different coordinates a_1, a_2, …, a_n. Each time when any of two cars drives through a flag, the speed of that car increases by 1 meter per second.

Find how long will it take for cars to meet (to reach the same coordinate). 

Input

The first line contains one integer t (1 ≤ t ≤ 10^4): the number of test cases.

The first line of each test case contains two integers n, l (1 ≤ n ≤ 10^5, 1 ≤ l ≤ 10^9): the number of flags and the length of the road.

The second line contains n integers a_1, a_2, …, a_n in the increasing order (1 ≤ a_1 < a_2 < … < a_n < l).

It is guaranteed that the sum of n among all test cases does not exceed 10^5.

Output

For each test case print a single real number: the time required for cars to meet.

Your answer will be considered correct, if its absolute or relative error does not exceed 10^{-6}. More formally, if your answer is a and jury's answer is b, your answer will be considered correct if \frac{|a-b|}{max{(1, b)}} ≤ 10^{-6}.

Example

Input


5
2 10
1 9
1 10
1
5 7
1 2 3 4 6
2 1000000000
413470354 982876160
9 478
1 10 25 33 239 445 453 468 477


Output


3.000000000000000
3.666666666666667
2.047619047619048
329737645.750000000000000
53.700000000000000

Note

In the first test case cars will meet in the coordinate 5.

The first car will be in the coordinate 1 in 1 second and after that its speed will increase by 1 and will be equal to 2 meters per second. After 2 more seconds it will be in the coordinate 5. So, it will be in the coordinate 5 in 3 seconds.

The second car will be in the coordinate 9 in 1 second and after that its speed will increase by 1 and will be equal to 2 meters per second. After 2 more seconds it will be in the coordinate 5. So, it will be in the coordinate 5 in 3 seconds.

In the second test case after 1 second the first car will be in the coordinate 1 and will have the speed equal to 2 meters per second, the second car will be in the coordinate 9 and will have the speed equal to 1 meter per second. So, they will meet after (9-1)/(2+1) = 8/3 seconds. So, the answer is equal to 1 + 8/3 = 11/3.
import java.util.*;
import java.lang.*;
import java.io.*;
import java.awt.*;

// U KNOW THAT IF THIS DAY WILL BE URS THEN NO ONE CAN DEFEAT U HERE................
//JUst keep faith in ur strengths .................................................. 


// ASCII = 48 + i ;// 2^28 = 268,435,456  > 2* 10^8 // log 10 base 2 = 3.3219 
// odd:: (x^2+1)/2 , (x^2-1)/2 ; x>=3// even:: (x^2/4)+1 ,(x^2/4)-1  x >=4 
// FOR ANY ODD NO N : N,N-1,N-2
//ALL ARE PAIRWISE COPRIME 
//THEIR COMBINED LCM IS PRODUCT OF ALL THESE NOS

// two consecutive odds are always coprime to each other
// two consecutive even have always gcd  = 2 ;

// Rectangle r = new Rectangle(int x , int y,int widht,int height) 
//Creates a rect. with bottom left cordinates as (x, y) and top right as ((x+width),(y+height))

//BY DEFAULT Priority Queue is MIN in nature in java
//to use as max , just push with negative sign and change sign after removal 

// We can make a sieve of max size 1e7 .(no time or space issue) 
// In 1e7 starting nos we have about 66*1e4 prime nos
// In 1e6 starting nos we have about 78,498 prime nos

 public class Main
{
     
    // static int[] arr = new int[100002] ; 
    // static int[] dp = new int[100002] ;  
    
     static PrintWriter out;
    
	static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		public FastReader(){
			br=new BufferedReader(new InputStreamReader(System.in));
			out=new PrintWriter(System.out);
		}
		String next(){
			while(st==null || !st.hasMoreElements()){
				try{
					st= new StringTokenizer(br.readLine());
				}
				catch (IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt(){
			return Integer.parseInt(next());
		}
		long nextLong(){
			return Long.parseLong(next());
		}
		double nextDouble(){
			return Double.parseDouble(next());
		}
		String nextLine(){
			String str = "";
			try{
				str=br.readLine();
			}
			catch(IOException e){
				e.printStackTrace();
			}
			return str;
		}
	}
	


////////////////////////////////////////////////////////////////////////////////////
 public static int countDigit(long n) 
    { 
        return (int)Math.floor(Math.log10(n) + 1); 
    } 

///////////////////////////////////////////////////////////////////////////////////////// 
 
 public static int sumOfDigits(long n)
 {
  
  if( n< 0)return -1 ;
  
  int sum = 0;
  
  while( n > 0)
  {
      sum = sum + (int)( n %10) ;
      
      n /= 10 ;
  }
     
  return sum ;  
 
 
 
 }
 
 //////////////////////////////////////////////////////////////////////////////////////////////////

public static long arraySum(int[] arr , int start , int end)
{
    long ans = 0 ;
    
    for(int i = start ; i <= end  ; i++)ans += arr[i] ;
    
    return ans  ;
}

/////////////////////////////////////////////////////////////////////////////////



////////////////////////////////////////////////////////////////////////////////

public static void swapArray(int[] arr , int start , int end)
{
    while(start < end)
    {
        int temp = arr[start] ;
        arr[start] = arr[end];
        arr[end] = temp;
        start++ ;end-- ;
    }
}


//////////////////////////////////////////////////////////////////////////////////

static long factorial(long a)
{
    if(a== 0L || a==1L)return 1L ;
    
    return a*factorial(a-1L) ;
}

///////////////////////////////////////////////////////////////////////////////


public static int[][] rotate(int[][] input){

int n =input.length;
int m = input[0].length ;
int[][] output = new int [m][n];

for (int i=0; i<n; i++)
	for (int j=0;j<m; j++)
		output [j][n-1-i] = input[i][j];
return output;
}
///////////////////////////////////////////////////////////////////////////////////////////////////////


/////////////////////////////////////////// ////////////////////////////////////////////////   

public static boolean isPowerOfTwo(long n) 
{ 
    if(n==0) 
    return false; 
  
if(((n ) & (n-1)) == 0 ) return true ;
else return false  ;

} 

/////////////////////////////////////////////////////////////////////////////////////

 
/////////////////////////////////////////////////////////////////////////////////// 

public static String reverse(String input)
{
  StringBuilder str  = new StringBuilder("") ;
   
    for(int i =input.length()-1 ; i >= 0  ; i-- )
    {
        str.append(input.charAt(i));
    }
    
return str.toString() ;
}
///////////////////////////////////////////////////////////////////////////////////////////



////////////////////////////////////////////////////////////////////////////////////////////////////
public static boolean isPossibleTriangle(int a  ,int b , int c)
{
      if( a + b > c && c+b > a && a +c > b)return true  ;
      else return false  ;
}


////////////////////////////////////////////////////////////////////////////////////////////
static long xnor(long num1, long num2) {
		if (num1 < num2) {
			long temp = num1;
			num1 = num2;
			num2 = temp;
		}
		num1 = togglebit(num1);
		return num1 ^ num2;
	}

	static long togglebit(long n) {
		if (n == 0)
			return 1;
		long i = n;
		n |= n >> 1;
		n |= n >> 2;
		n |= n >> 4;
		n |= n >> 8;
		n |= n >> 16;
		return i ^ n;
	}

///////////////////////////////////////////////////////////////////////////////////////////////

public static int xorOfFirstN(int n)
{
 
 
 if( n % 4 ==0)return n ;
 
 else if( n % 4 == 1)return 1 ;
 
 else if( n % 4 == 2)return n+1 ;
 
 else return 0 ;
 
    
}
//////////////////////////////////////////////////////////////////////////////////////////////

public static int gcd(int a, int b )
{

if(b==0)return a ;

else return gcd(b,a%b) ; 

}


public static long gcd(long a, long b )
{

if(b==0)return a ;

else return gcd(b,a%b) ; 

}

////////////////////////////////////////////////////////////////////////////////////

public static int lcm(int a, int b ,int c , int d )
{

int temp = lcm(a,b , c) ;


 
 int ans = lcm(temp ,d ) ;

return ans  ;


}

///////////////////////////////////////////////////////////////////////////////////////////

public static int lcm(int a, int b ,int c )
{

int temp = lcm(a,b) ;

int ans =  lcm(temp ,c) ;

return ans  ;


}

////////////////////////////////////////////////////////////////////////////////////////
    
public static int lcm(int a , int b )
{

int gc = gcd(a,b);

return (a/gc)*b ;
}


public static long lcm(long a , long b )
{

long gc = gcd(a,b);

return (a/gc)*b;
}


///////////////////////////////////////////////////////////////////////////////////////////
static boolean isPrime(long n)
{
      if(n==1)
      {
            return false  ;
      }
      
      boolean ans =  true  ;
      
      for(long i = 2L; i*i <= n ;i++)
      {
            if(n% i ==0)
            {
                  ans = false  ;break ;
            }
      }
      
      
      return ans  ;
} 

static boolean isPrime(int n)
{
      if(n==1)
      {
            return false  ;
      }
      
      boolean ans =  true  ;
      
      for(int i = 2; i*i <= n ;i++)
      {
            if(n% i ==0)
            {
                  ans = false  ;break ;
            }
      }
      
      
      return ans  ;
}      


///////////////////////////////////////////////////////////////////////////

static int sieve =  1000000 ;

 
static boolean[] prime =  new boolean[sieve + 1] ;

public static void sieveOfEratosthenes() 
    { 
        // FALSE == prime
        
        // TRUE ==  COMPOSITE
        
        // FALSE== 1
       
        
        // time complexity = 0(NlogLogN)== o(N)
        
        // gives prime nos bw 1 to N
        
        for(int i = 4; i<= sieve ; i++)
        {
            prime[i] = true  ;
            i++ ;
        }
        
        for(int p = 3; p*p <= sieve; p++) 
        { 
           
            if(prime[p] == false) 
            { 
                
                for(int i = p*p; i <= sieve; i += p) 
                    prime[i] = true; 
            } 
            
            p++ ;
        } 
          
       
       
    
    } 
 
///////////////////////////////////////////////////////////////////////////////////


public static void sortD(int[] arr , int s  , int e)
{
      sort(arr ,s , e) ;
      
      int i =s ; int j = e  ;
      
      while( i < j)
      {
            int temp = arr[i] ;
            arr[i] =arr[j] ;
            arr[j] = temp ;
            i++ ; j-- ;
      }
      
      
      
      return ;
}


/////////////////////////////////////////////////////////////////////////////////////////

public static long countSubarraysSumToK(long[] arr ,long sum )
    {
      HashMap<Long,Long> map = new HashMap<>() ;
       
      int n = arr.length ;
       
      long prefixsum = 0 ;
       
      long count = 0L ;
      for(int i  = 0; i < n ; i++)
      {
          prefixsum  = prefixsum +  arr[i] ;
           
          if(sum == prefixsum)count = count+1 ;
           
          if(map.containsKey(prefixsum -sum))
          {
              count = count + map.get(prefixsum -sum) ;
          }
          
          
          if(map.containsKey(prefixsum ))
          {
              map.put(prefixsum , map.get(prefixsum) +1 );
          }
          
          else{
              map.put(prefixsum , 1L );
          }
           
           
      }
        
        
        
      return count  ;  
        
    }

///////////////////////////////////////////////////////////////////////////////////////////////


// KMP ALGORITHM : TIME COMPL:O(N+M) 
// FINDS THE OCCURENCES OF PATTERN AS A  SUBSTRING IN STRING
//RETURN THE ARRAYLIST OF INDEXES 
// IF SIZE OF LIST IS ZERO MEANS PATTERN IS NOT PRESENT IN STRING


public static ArrayList<Integer> kmpAlgorithm(String str , String pat)
     {
        ArrayList<Integer> list =new ArrayList<>();
        
        int n = str.length() ;
        int m = pat.length() ;
        
        String q = pat + "#" + str ;
        
        int[] lps  =new int[n+m+1] ;
        
         longestPefixSuffix(lps, q,(n+m+1)) ;
         
         
         for(int i =m+1 ; i < (n+m+1) ; i++ )
         {
             if(lps[i] == m)
             {
                 list.add(i-2*m) ;
             }
         }
        
        return list ; 
         
        
    }
    

public static void longestPefixSuffix(int[] lps ,String str , int n)
    {
        lps[0] =  0 ;
        
        for(int i = 1  ; i<= n-1; i++)
        {
          int l = lps[i-1] ;
           
          while( l > 0 && str.charAt(i) != str.charAt(l))
          {
              l = lps[l-1] ;
          }
           
          if(str.charAt(i) == str.charAt(l))
          {
              l++ ;
          }
            
           
          lps[i] = l ; 
        }
        
    }
    
     

///////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////

 // CALCULATE TOTIENT Fn FOR ALL VALUES FROM 1 TO n
    // TOTIENT(N) = count of nos less than n and grater than 1 whose gcd with n is 1 
    // or n and the no  will be coprime in nature
    //time : O(n*(log(logn)))
    
    public static void eulerTotientFunction(int[] arr ,int n )
    {
      
      for(int i = 1; i <= n  ;i++)arr[i] =i  ;
      
      
      for(int i= 2 ; i<= n ;i++)
      {
          if(arr[i] == i)
          {
              arr[i] =i-1 ;
              
              for(int j =2*i ; j<= n  ; j+= i )
              {
                  arr[j] = (arr[j]*(i-1))/i ;
              }
              
          }
      }
        
      return  ;  
        
    }
	
/////////////////////////////////////////////////////////////////////////////////////////////
public static long nCr(int n,int k)
{
    long ans=1L;
    k=k>n-k?n-k:k;
    int j=1;
    for(;j<=k;j++,n--)
    {
        if(n%j==0)
        {
            ans*=n/j;
        }else
        if(ans%j==0)
        {
            ans=ans/j*n;
        }else
        {
            ans=(ans*n)/j;
        }
    }
    return ans;
}

///////////////////////////////////////////////////////////////////////////////////////////

public static ArrayList<Integer> allFactors(int n)
{   
      ArrayList<Integer> list = new ArrayList<>() ;
      
    for(int i = 1; i*i <= n ;i++)
    {
          if( n % i == 0)
          {
              if(i*i == n)
              {
                    list.add(i) ;
              }
              else{
                    list.add(i) ;
                    list.add(n/i) ;
                    
              }
          }
    }
      
     return list ; 
      
      
}


public static ArrayList<Long> allFactors(long n)
{   
      ArrayList<Long> list = new ArrayList<>() ;
      
    for(long i = 1L; i*i <= n ;i++)
    {
          if( n % i == 0)
          {
              if(i*i == n)
              {
                    list.add(i) ;
              }
              else{
                    list.add(i) ;
                    list.add(n/i) ;
                    
              }
          }
    }
      
     return list ; 
      
      
}
////////////////////////////////////////////////////////////////////////////////////////////////////

  static final int MAXN = 1000001; 
       
    
    static int spf[] = new int[MAXN]; 
   
    static void sieve() 
    { 
        spf[1] = 1; 
        for (int i=2; i<MAXN; i++) 
       
           
            spf[i] = i; 
       
       
        for (int i=4; i<MAXN; i+=2) 
            spf[i] = 2; 
       
        for (int i=3; i*i<MAXN; i++) 
        { 
           
            if (spf[i] == i) 
            { 
               
                for (int j=i*i; j<MAXN; j+=i) 
       
                    if (spf[j]==j) 
                        spf[j] = i; 
            } 
        } 
    } 
       

    static ArrayList<Integer> getPrimeFactorization(int x) 
    { 
        ArrayList<Integer> ret = new ArrayList<Integer>(); 
        while (x != 1) 
        { 
            ret.add(spf[x]); 
            x = x / spf[x]; 
        } 
        return ret; 
    } 
       
 //////////////////////////////////////////////////////////////////////////////////////////////////
 //////////////////////////////////////////////////////////////////////////////////////////////////
   
public static void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int[n1];
        int R[] = new int[n2];
 
       //Copy data to temp arrays
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
  public static void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

public static void sort(long arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }


public static void merge(long arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        long L[] = new long[n1];
        long R[] = new long[n2];
 
        //Copy data to temp arrays
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 

 /////////////////////////////////////////////////////////////////////////////////////////
 
	public static long knapsack(int[] weight,long value[],int maxWeight){

        
        int n=  value.length ;
        
	
	//dp[i] stores the profit with KnapSack capacity "i" 
  long []dp = new long[maxWeight+1]; 
      
    //initially profit with 0 to W KnapSack capacity is 0 
    Arrays.fill(dp, 0); 
  
    // iterate through all items 
    for(int i=0; i < n; i++)  
      
        //traverse dp array from right to left 
        for(int j = maxWeight; j >= weight[i]; j--) 
            dp[j] = Math.max(dp[j] , value[i] + dp[j - weight[i]]); 
              
    /*above line finds out maximum of dp[j](excluding ith element value) 
    and val[i] + dp[j-wt[i]] (including ith element value and the 
    profit with "KnapSack capacity - ith element weight") */
    return dp[maxWeight]; 
	}

///////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////


// to return max sum of any subarray in given array
public static long kadanesAlgorithm(long[] arr)
{
    
    if(arr.length == 0)return  0 ;
    
      long[] dp = new long[arr.length] ;
      
      dp[0] = arr[0] ;
      long max =  dp[0] ;
      
      
      for(int i = 1; i <  arr.length ; i++)
      {
            if(dp[i-1] > 0)
            {
                  dp[i] = dp[i-1] + arr[i] ;
            }
            else{
                  dp[i] = arr[i] ;
            }
            
            if(dp[i] >  max)max = dp[i] ;
            
      }
      
      return max  ;
      
}
/////////////////////////////////////////////////////////////////////////////////////////////
public static long kadanesAlgorithm(int[] arr)
{
     if(arr.length == 0)return  0 ;
    
      long[] dp = new long[arr.length] ;
      
      dp[0] = arr[0] ;
      long max =  dp[0] ;
      
      
      for(int i = 1; i <  arr.length ; i++)
      {
            if(dp[i-1] > 0)
            {
                  dp[i] = dp[i-1] + arr[i] ;
            }
            else{
                  dp[i] = arr[i] ;
            }
            
            if(dp[i] >  max)max = dp[i] ;
            
      }
      
      return max  ;
      
}

      
///////////////////////////////////////////////////////////////////////////////////////



/////////////////////////////////////////////////////////////////////////////////////////



//////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////

//TO GENERATE ALL(DUPLICATE ALSO EXIST) PERMUTATIONS OF A STRING


// JUST CALL generatePermutation( str,  start,  end) start :inclusive ,end : exclusive 

//Function for swapping the characters at position I with character at position j  
    public static String swapString(String a, int i, int j) {  
        char[] b =a.toCharArray();  
        char ch;  
        ch = b[i];  
        b[i] = b[j];  
        b[j] = ch;  
        return String.valueOf(b);  
    }  
      
//Function for generating different permutations of the string  
    public static void generatePermutation(String str, int start, int end)  
    {  
        //Prints the permutations  
        if (start == end-1)  
            System.out.println(str);  
        else  
        {  
            for (int i = start; i < end; i++)  
            {  
                //Swapping the string by fixing a character  
                str = swapString(str,start,i);  
                //Recursively calling function generatePermutation() for rest of the characters   
                generatePermutation(str,start+1,end);  
                //Backtracking and swapping the characters again.  
                str = swapString(str,start,i);  
            }  
        }  
    }  



////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////

public static long factMod(long n, long mod) {
    if (n <= 1) return 1;
    long ans = 1;
    for (int i = 1; i <= n; i++) {
      ans = (ans * i) % mod;
    }
    return ans;
  }


/////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////

public static long power(int a  ,int b)
    {
        //time comp : o(logn) 
        
        long x = (long)(a) ;
         long n = (long)(b) ;
        
        if(n==0)return 1 ;
        if(n==1)return x;
        
        long ans =1L  ;
       
      while(n>0)
      {
          if(n % 2 ==1)
          {
              ans = ans *x ;
          }
           
          n = n/2L ;
           
          x =  x*x ;
           
      }
       
      return ans ;
    }
    
    public static long power(long a  ,long b)
    {
        //time comp : o(logn) 
        
        long x = (a) ;
         long n = (b) ;
        
        if(n==0)return 1L ;
        if(n==1)return x;
        
        long ans =1L  ;
       
      while(n>0)
      {
          if(n % 2 ==1)
          {
              ans = ans *x ;
          }
           
          n = n/2L ;
           
          x =  x*x ;
           
      }
       
      return ans ;
    }

    
    
    

////////////////////////////////////////////////////////////////////////////////////////////////////
public static long powerMod(long x, long n, long mod) {
    //time comp : o(logn)
    
    if(n==0)return 1L ;
        if(n==1)return x;
        
    
    long ans = 1;
    while (n > 0) {
      if (n % 2 == 1) ans = (ans * x) % mod;
      x = (x * x) % mod;
      n /= 2;
    }
    return ans;
  }
 
//////////////////////////////////////////////////////////////////////////////////

/////////////////////////////////////////////////////////////////////////////////

/*
lowerBound - finds largest element equal or less than value paased
upperBound - finds smallest element equal or more than value passed

if not present return -1;

*/

public static long lowerBound(long[] arr,long k)
	{
		long ans=-1;
		
		int start=0;
		int end=arr.length-1;
		
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]<=k)
			{
				ans=arr[mid];
				start=mid+1;
			}
			else
			{
				end=mid-1;
			}
			
		}
		
		return ans;
		
	}
	
	public static int lowerBound(int[] arr,int k)
	{
		int ans=-1;
		
		int start=0;
		int end=arr.length-1;
		
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]<=k)
			{
				ans=arr[mid];
				start=mid+1;
			}
			else
			{
				end=mid-1;
			}
			
		}
		
		return ans;
		
	}
	
	
	public static long upperBound(long[] arr,long k)
	{
		long ans=-1;
		
		int start=0;
		int end=arr.length-1;
		
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]>=k)
			{
				ans=arr[mid];
				end=mid-1;
			}
			else
			{
				start=mid+1;
			}
			
		}
		
		return ans;
	}
	
	
	public static int upperBound(int[] arr,int k)
	{
		int ans=-1;
		
		int start=0;
		int end=arr.length-1;
		
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]>=k)
			{
				ans=arr[mid];
				end=mid-1;
			}
			else
			{
				start=mid+1;
			}
			
		}
		
		return ans;
	}
	

//////////////////////////////////////////////////////////////////////////////////////////

public static void printArray(int[] arr , int si ,int ei)
{
    for(int i = si  ; i  <= ei ; i++)
    {
        out.print(arr[i] +" ") ;
    }
    
}

public static void printArrayln(int[] arr , int si ,int ei)
{
    for(int i = si  ; i  <= ei ; i++)
    {
        out.print(arr[i] +" ") ;
    }
    out.println() ;
}


public static void printLArray(long[] arr , int si , int ei)
{
    for(int i = si ; i  <= ei ; i++)
    {
        out.print(arr[i] +" ") ;
    }
   
}




public static void printLArrayln(long[] arr , int si , int ei)
{
    for(int i = si ; i  <= ei ; i++)
    {
        out.print(arr[i] +" ") ;
    }
    out.println() ;
   
}

public static void printtwodArray(int[][] ans)
{
    for(int i = 0; i< ans.length ; i++)
    {
        for(int j  = 0 ; j <  ans[0].length ; j++)out.print(ans[i][j] +" ");
        out.println() ;
    }
    out.println() ;
   
}

  
   static long modPow(long a, long x, long p) {
    //calculates a^x mod p in logarithmic time.
    
    a = a % p ;
    
    if(a == 0)return 0L ;
    
    
    long res = 1L;
    while(x > 0) {
        if( x % 2 != 0) {
            res = (res * a) % p;
        }
        a = (a * a) % p;
        x =x/2;
    }
    return res;
}
 
 
 
  static long modInverse(long a, long p) {
    //calculates the modular multiplicative of a mod p.
    //(assuming p is prime).
    return modPow(a, p-2, p);
}
 
 static long[] factorial =  new long[1000001] ;
 
 static void modfac(long mod)
 {
     factorial[0]=1L ; factorial[1]=1L ;
     
     for(int i = 2; i<= 1000000 ;i++)
     {
         factorial[i] = factorial[i-1] *(long)(i) ;
          factorial[i] = factorial[i] % mod ;
     }
    
     
 }
 

 
 
 
static long modBinomial(long n, long r, long p) {
// calculates C(n,r) mod p (assuming p is prime).
 
  if(n < r) return 0L ; 
 
    long num = factorial[(int)(n)] ;
    
    long den = (factorial[(int)(r)]*factorial[(int)(n-r)]) %  p ;
    
    
    long ans  = num*(modInverse(den,p)) ;
    
    ans = ans % p ;
    
    return ans  ;
 
    
}
 
 
 static void update(int val , long[] bit ,int n)
 {
     for( ; val <= n ; val += (val &(-val)) )
     {
         bit[val]++ ;
     }
    
     
 }
 
 
 static long query(int val , long[] bit , int n)
 {
     long ans = 0L; 
     for( ; val >=1 ; val-=(val&(-val))  )ans += bit[val];
     
     return ans ;
 }



static int countSetBits(long n) 
    { 
        int count = 0; 
        while (n > 0) { 
            n = (n) & (n - 1L); 
            count++; 
        } 
        return count; 
    } 


static int abs(int x)
{
    if(x < 0)x = -1*x ;
    
    return x ;
}


static long abs(long x)
{
    if(x < 0)x = -1L*x ;
    
    return x ;

}

////////////////////////////////////////////////////////////////////////////////////////////////
static void p(int val)
{
    out.print(val) ;
}

static void p()
{
    out.print(" ") ;
}

static void pln(int val)
{
    out.println(val) ;
}

static void pln()
{
    out.println() ;
}


static void p(long val)
{
    out.print(val) ;
}



static void pln(long val)
{
    out.println(val) ;
}


static void yes()
{
    out.println("YES") ;
}
static void no()
{
    out.println("NO") ;
}

////////////////////////////////////////////////////////////////////////////////////////////

 // calculate total no of nos greater than or equal to key in  sorted array arr
 
static  int bs(int[] arr, int s ,int e  ,int key)
{
    if( s> e)return 0 ;
    
     int mid = (s+e)/2 ;
        
        if(arr[mid] <key)
        {
            return bs(arr ,mid+1,e , key) ;
            
        }
        
        
        else{
            
             return bs(arr ,s ,mid-1, key)  + e-mid+1;
            
            
        }
}
 
// static ArrayList<Integer>[] adj ;
// static int mod= 1000000007 ;


 



//////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////

public static void solve()
{
FastReader scn = new FastReader() ;

//DEBUGGING -
// 1 - I/O - CHECK PRINTING OF NEXT LINE , READ  INPUT , EXTRA LINE PRINT.
//2-OVERFLOW ERROR.
//3 -STATIC VARIABLE INITIALIZATION (INT MAINLIY)
//4-DIVISION ERROR(FLOOR/CEIL) or DIVIDE BY ZERO
//5-MOD -(MOD BY 1 OR 0 )
// DOUBLE LOOP/TRIPLE LOOP BREAK (SEE IF DO NOT BREAK ALL THE LOOPS IF U GET ANS EARLIER)


//Scanner scn = new Scanner(System.in);
//int[] store = {2 ,3, 5 , 7  ,11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 } ;

// product of first 11 prime nos is greater than 10 ^ 12;
//sieve() ;
//ArrayList<Integer> arr[] = new ArrayList[n] ;
ArrayList<Integer> list = new ArrayList<>() ;
ArrayList<Long> lista = new ArrayList<>() ;
ArrayList<Long> listb = new ArrayList<>() ;
// ArrayList<Integer> lista = new ArrayList<>() ;
// ArrayList<Integer> listb = new ArrayList<>() ;
//ArrayList<String> lists = new ArrayList<>() ;

HashMap<Integer,Integer> map = new HashMap<>() ;
//HashMap<Long,Long> map = new HashMap<>() ;
HashMap<Integer,Integer> mapx  = new HashMap<>() ;
HashMap<Integer,Integer> mapy  = new HashMap<>() ;
//HashMap<String,Integer> maps = new HashMap<>() ;
//HashMap<Integer,Boolean> mapb = new HashMap<>() ;
//HashMap<Point,Integer> point = new HashMap<>() ; 

 Set<Integer> set = new HashSet<>() ;
 Set<Integer> setx = new HashSet<>() ;
 Set<Integer> sety = new HashSet<>() ;

StringBuilder sb =new StringBuilder("") ;

//Collections.sort(list);

//if(map.containsKey(arr[i]))map.put(arr[i] , map.get(arr[i]) +1 ) ;
//else map.put(arr[i],1) ;

// if(map.containsKey(temp))map.put(temp , map.get(temp) +1 ) ;
// else map.put(temp,1) ;

//int bit  =Integer.bitCount(n);
// gives total no of set bits in n;

// Arrays.sort(arr, new Comparator<Pair>() {
// 			@Override
// 			public int compare(Pair a, Pair b) {
// 				if (a.first != b.first) {
// 					return a.first - b.first; // for increasing order of first
// 				}
// 				return a.second - b.second ; //if first is same then sort on second basis
// 			}
// 		});


int testcase = 1;
 testcase = scn.nextInt() ;
for(int testcases =1  ; testcases <= testcase ;testcases++)
{
    
 //if(map.containsKey(arr[i]))map.put(arr[i],map.get(arr[i])+1) ;else map.put(arr[i],1) ;
 //if(map.containsKey(temp))map.put(temp,map.get(temp)+1) ;else map.put(temp,1) ;
 
 //adj = new ArrayList[n] ;

// for(int i = 0; i< n; i++)
// {
//     adj[i] = new ArrayList<Integer>();
// }

// long n = scn.nextLong() ;
//String s = scn.next() ;

int n= scn.nextInt() ;double  l= scn.nextDouble() ;
double[] arr= new double[n+2] ; 
arr[0] = 0 ; arr[n+1] = l ;

for(int i=1; i <= n;i++)
{
    arr[i]= scn.nextDouble();
}

double[] lt = new double[n+2] ;
double[] rt = new double[n+2] ;

lt[0] = 0.0 ;rt[n+1] = 0.0 ;


for(int i = 1 ; i <= n+1 ; i++ )
{
    double d = arr[i] - arr[i-1] ;    
    
    double time  = d/(i) ;
    
    lt[i] = lt[i-1] + (time);

}


for(int i = n ; i >=  0 ; i-- )
{
    double d = arr[i+1] - arr[i] ;    
    
    double time  = d/(n+1-i) ;
    
    rt[i] = rt[i+1] + (time);

}

int in = -1  ;

for(int i = 0 ; i <= n+1 ;i++)
{
    if(lt[i] >=rt[i])
    {
        
        in = i ;break ;
    }
    
}

if(lt[in] == rt[in])out.println(lt[in]) ;
else{
    
    double d= arr[in] - arr[in-1] ;
    
    
  double t1 = lt[in-1] ; double t2 = rt[in] ;
    
    double time ;
    
    if(t1 < t2)
    {
        
        
        time  = t2 ;
        
        d=d-((in)*(t2-t1)) ;
        
        
    }
    else{
        // t1 > t2
        
        time =t1  ;
          
        d=d-((n+1-(in-1))*(t1-t2)) ;
        
    }
    
    
    double ans = time + (d)/(in + n+1 -in +1)  ;
    
    out.println(ans) ;
    
}




//out.println(ans) ;
//out.println(ans+" "+in) ;

//out.println("Case #" + testcases + ": "  + ans ) ;
//out.println("@") ;
set.clear() ;
sb.delete(0 , sb.length()) ;
list.clear() ;lista.clear() ;listb.clear() ;
map.clear() ;
mapx.clear() ;
mapy.clear() ;
setx.clear() ;sety.clear() ;

} // test case  end loop


out.flush() ;  
} // solve fn ends


public static void main (String[] args) throws java.lang.Exception
{
  

solve() ;
      
}


}
  




 class Pair 
{
  int first ;
  
int second  ;
  
  public  Pair(int x, int y)
  {
      this.first = x ;this.second = y ;
  }
  
  @Override
 public boolean equals(Object obj)
 {
     if(obj == this)return true  ;
     
     if(obj == null)return false  ;
     
     if(this.getClass() != obj.getClass())
     {
         return false ;
     }
     
     
     Pair other  = (Pair)(obj) ;
     
     if(this.first != other.first)return false  ;
     
     
     if(this.second != other.second)return false  ;
     
     return true  ;
     
 }
    
    @Override
    
    public int hashCode()
    {
        return this.first^this.second  ;
    }
      
@Override
	public String toString() {
	
	String ans = "" ;
	ans  += this.first ;
	ans += " ";
	ans += this.second ;
	
	return ans  ;
	}


}



