Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.

For example, if s =  baba and k=2, Phoenix may distribute the letters of his string in many ways, such as: 

  * ba and ba
  * a and abb
  * ab and ab
  * aa and bb



But these ways are invalid: 

  * baa and ba
  * b and ba
  * baba and empty string (a_i should be non-empty) 



Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).

String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.

The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.

The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.

It is guaranteed that the sum of n over all test cases is ≤ 10^5.

Output

Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.

Example

Input


6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix


Output


ab
abbc
b
aa
x
ehinopx

Note

In the first test case, one optimal solution is to distribute baba into ab and ab. 

In the second test case, one optimal solution is to distribute baacb into abbc and a.

In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.

In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.

In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.

In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
import java.util.*;
import java.lang.*;
import java.io.*;

// SHIVAM GUPTA :
//NSIT
//decoder_1671
//BEING PERFECTIONIST IS NOT AN OPTION



// ASCII = 48 + i ;

// SHIVAM GUPTA :


/* Name of the class has to be "Main" only if the class is public. */



public class Main
{
     
     
     
    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  min(int a ,int b , int c, int d)
{
       
       
       int[] arr = new int[4] ;
       arr[0] = a;
       arr[1] = b ;
       arr[2] = c;
       arr[3] = d;
       Arrays.sort(arr) ;
       
       return arr[0];
       
}
 
 /////////////////////////////////////////////////////////////////////////////
 
 
public static int  max(int a ,int b , int c, int d)
{
       
       
       int[] arr = new int[4] ;
       arr[0] = a;
       arr[1] = b ;
       arr[2] = c;
       arr[3] = d;
       Arrays.sort(arr) ;
       
       return arr[3];
       
}
 
/////////////////////////////////////////////////////////////////////////////////// 
static int sieve =  1000000 ;

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

static int[] dp =  new int[10000001] ;

 
public static void sieveOfEratosthenes() 
    { 
        // FALSE == prime
        
        // TRUE ==  COMPOSITE
        
        // FALSE== 1
       
        for(int i=0;i< sieve + 1;i++) 
            prime[i] = false; 
          
        for(int p = 2; p*p <= sieve; p++) 
        { 
           
            if(prime[p] == false) 
            { 
                
                for(int i = p*p; i <= sieve; i += p) 
                    prime[i] = true; 
            } 
        } 
          
       
       
    
    } 
 
///////////////////////////////////////////////////////////////////////////////////


public static String reverse(String input)
{
    String op = "" ;
    
    for(int i = 0; i < input.length() ; i++ )
    {
        op = input.charAt(i)+  op ; 
    }
    
return op ;
}


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


public static void sortD(int[] arr)
{
      sort(arr ,0 , arr.length-1) ;
      
      int i =0 ; int j = arr.length -1  ;
      
      while( i < j)
      {
            int temp = arr[i] ;
            arr[i] =arr[j] ;
            arr[j] = temp ;
            i++ ; j-- ;
      }
      
      
      
      return ;
}

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

public static boolean sameParity(long a  ,long b )
{
      
      if(a%2 == b %2)
      {
            return true  ;
      }
      else{
            return false  ;
      }
}


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

public static boolean sameParity(int a  ,int b )
{
      
      if(a%2 == b %2)
      {
            return true  ;
      }
      else{
            return false  ;
      }
}

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

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

/////////////////////////////////////////////////////////////////////
    
public static int gcd(int a, int 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*b)/gc ;




}

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

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  ;
      
      
}
//////////////////////////////////////////////////////////////////////////////////////////
public static long nCr(int n,int k)
{
    long ans=1;
    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 boolean isPowerOfTwo(int n) 
{ 
    if(n==0) 
    return false; 
  
if(((n ) & (n-1)) == 0 ) return true ;
else return false  ;

} 
   
////////////////////////////////////////////////////////////////////////////////////   
   
    public static int countDigit(long n) 
    { 
        return (int)Math.floor(Math.log10(n) + 1); 
    } 
    
 ///////////////////////////////////////////////////////////////////////////////////////// 
  
   static final int MAXN = 100001; 
       
    
    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; 
            } 
        } 
    } 
       
// The above code works well for n upto the order of 10^7.
// Beyond this we will face memory issues.

// Time Complexity: The precomputation for smallest prime factor is done in O(n log log n)
// using sieve.
// Where as in the calculation step we are dividing the number every time by 
// the smallest prime number till it becomes 1.
// So, let’s consider a worst case in which every time the SPF is 2 .
// Therefore will have log n division steps.


// Hence, We can say that our Time Complexity will be O(log n) in worst case.


    static Vector<Integer> getFactorization(int x) 
    { 
        Vector<Integer> ret = new Vector<>(); 
        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 int knapsack(int[] weight,int value[],int maxWeight){

        
        int n=  value.length ;
        
	
	//dp[i] stores the profit with KnapSack capacity "i" 
    int []dp = new int[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]; 
	}

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

public static long gcd(long a  , long b)
{
    if(b==0)return a  ;
    
    return gcd(b , a % b) ;
    
}

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

public static long gcdOfArray(long[] arr)
{
    long ans = arr[0] ;
    
    for(int i = 1; i < arr.length ; i++)
    {
        ans = gcd(ans,arr[i]) ;
    }
    
    return ans  ; 
    
}

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


// to return max sum of any subarray in given array
public static long kadanesAlgorithm(long[] arr)
{
      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)
{
      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 binarySerachGreater(int[] arr , int start , int end , int val)
{
      
      // fing total no of elements strictly grater than val in sorted array arr 
      
      
      if(start >  end)return  0 ; //Base case
      
      
      int mid = (start + end)/2  ;
      
      if(arr[mid] <=val)
      {
           return binarySerachGreater(arr,mid+1, end ,val) ; 
            
      }
      else{
            
         return binarySerachGreater(arr,start , mid -1,val) + end-mid+1 ;    
            
      }
      
      
}
///////////////////////////////////////////////////////////////////////////////

// Arrays.sort(arr, new Comparator<Pair>() {
// 			@Override
// 			public int compare(Pair first, Pa second) {
// 				if (first.getAge() != second.getAge()) {
// 					return first.getAge() - second.getAge();
// 				}
// 				return first.getName().compareTo(second.getName());
// 			}
// 		});

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


public static void main (String[] args) throws java.lang.Exception
{
  FastReader scn = new FastReader() ;
  
  int t = scn.nextInt() ;
   Set<Character> set = new HashSet<>() ;
  
  while(t-->0)
  {
    
int n = scn.nextInt() ;int k = scn.nextInt() ;

String str = scn.next();

 char[] arr = str.toCharArray();
 
 Arrays.sort(arr) ;

if(arr[k-1] != arr[0] )out.print(arr[k-1]);

else if(k == n)
{
    out.print(arr[0]) ;
}
   
else if( arr[k] == arr[n-1])
{
    out.print(arr[0]) ;
    
    int count  =  (n-k)/k ;
    
    if((n-k) % k !=0 )count++ ;
    
    while(count-- > 0)
    {
        out.print(arr[k]) ;
    }
    
}
   
  
 else{      
       
       out.print(arr[0]) ;
       
       for(int i =  k ; i <  n; i++)
       {
           out.print(arr[i]) ;
       }
       
       
  
}

out.println() ; set.clear() ;
}

out.flush() ;
      
}


}
  
 class Pair 
{
  int first ;
  
  int second  ;
      
@Override



	public String toString() {
	
	String ans = "" ;
	ans  += this.first ;
	ans += " ";
	ans += this.second ;
	
	return ans  ;
	}



}



 