Recently, on the course of algorithms and data structures, Valeriy learned how to use a deque. He built a deque filled with n elements. The i-th element is a_i (i = 1, 2, …, n). He gradually takes the first two leftmost elements from the deque (let's call them A and B, respectively), and then does the following: if A > B, he writes A to the beginning and writes B to the end of the deque, otherwise, he writes to the beginning B, and A writes to the end of the deque. We call this sequence of actions an operation.

For example, if deque was [2, 3, 4, 5, 1], on the operation he will write B=3 to the beginning and A=2 to the end, so he will get [3, 4, 5, 1, 2].

The teacher of the course, seeing Valeriy, who was passionate about his work, approached him and gave him q queries. Each query consists of the singular number m_j (j = 1, 2, …, q). It is required for each query to answer which two elements he will pull out on the m_j-th operation.

Note that the queries are independent and for each query the numbers A and B should be printed in the order in which they will be pulled out of the deque.

Deque is a data structure representing a list of elements where insertion of new elements or deletion of existing elements can be made from both sides.

Input

The first line contains two integers n and q (2 ≤ n ≤ 10^5, 0 ≤ q ≤ 3 ⋅ 10^5) — the number of elements in the deque and the number of queries. The second line contains n integers a_1, a_2, ..., a_n, where a_i (0 ≤ a_i ≤ 10^9) — the deque element in i-th position. The next q lines contain one number each, meaning m_j (1 ≤ m_j ≤ 10^{18}).

Output

For each teacher's query, output two numbers A and B — the numbers that Valeriy pulls out of the deque for the m_j-th operation.

Examples

Input


5 3
1 2 3 4 5
1
2
10


Output


1 2
2 3
5 2


Input


2 0
0 0


Output

Note

Consider all 10 steps for the first test in detail:
  1. [1, 2, 3, 4, 5] — on the first operation, A and B are 1 and 2, respectively.

So, 2 we write to the beginning of the deque, and 1 — to the end.

We get the following status of the deque: [2, 3, 4, 5, 1].

  2. [2, 3, 4, 5, 1] ⇒ A = 2, B = 3.
  3. [3, 4, 5, 1, 2]
  4. [4, 5, 1, 2, 3]
  5. [5, 1, 2, 3, 4]
  6. [5, 2, 3, 4, 1]
  7. [5, 3, 4, 1, 2]
  8. [5, 4, 1, 2, 3]
  9. [5, 1, 2, 3, 4]
  10. [5, 2, 3, 4, 1] ⇒ A = 5, B = 2. 
import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		FastReader sc=new FastReader();
		int n=sc.I();
		int q=sc.I();
	   long[] a=new long[n];
	   int m=0;
	   for(int i=0;i<n;i++){
		   a[i]=sc.L();
		   if(a[m]<a[i])
		   m=i;}
		   Queue<Long> r=new LinkedList<>();
		   long t=a[0];
		   for(int i=1;i<n;i++)
		   r.add(a[i]);
		   long[][] s=new long[m+1][2];
		   s[0][0]=t;
		   s[0][1]=a[1];
		   for(int i=1;i<=m;i++)
		   {
		       long x=r.poll();
		       //long h=x+t;
		       if(x>t)
		       {
		           r.add(t);
		           t=x;
		           s[i][1]=r.peek();
		       }
		       else{
		       s[i][1]=r.peek();
		       r.add(x);
		       }
		       s[i][0]=t;
		       
		   }
		   ArrayList<Long> d=new ArrayList<>();
		   while(r.size()>0)
		   d.add(r.poll());
		   while(q-->0)
		   {
		       long w=sc.L();
		       if(w<=(long)m)
		       System.out.println(s[(int)w-1][0]+" "+s[(int)w-1][1]);
		       else{
		           w-=m;
		           w=w%(n-1);
		           if(w==0)
		           w=n-1;
		           System.out.println(t+" "+d.get((int)w-1));
		       }
		   }
	   
	 }
	 
	 static long power(long x, long y, long p) 
    { 
        // Initialize result 
        long res = 1;      
         
        // Update x if it is more   
        // than or equal to p 
        x = x % p;  
      
        while (y > 0) 
        { 
            // If y is odd, multiply x 
            // with result 
            if((y & 1)==1) 
                res = (res * x) % p; 
      
            // y must be even now 
            // y = y / 2 
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
    } 
	  static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int I() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long L() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double D() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
    static int gcd(int a,int b)
		{
		    if(a%b==0)
		    return b;
		    return gcd(b,a%b);
		}
	static float power(float x, int y) 
    { 
        float temp; 
        if( y == 0) 
            return 1; 
        temp = power(x, y/2);  
          
        if (y%2 == 0) 
            return temp*temp; 
        else
        { 
            if(y > 0) 
                return x * temp * temp; 
            else
                return (temp * temp) / x; 
        } 
    } 
    static long pow(int a,int b)
    {
        long result=1;
        if(b==0)
        return 1;
        long x=a;
        while(b>0)
        {
            if(b%2!=0)
            result*=x;
            
            x=x*x;
            b=b/2;
        }
        return result;
    }
    
    static ArrayList<Integer> sieveOfEratosthenes(int n) 
    { 
        ArrayList<Integer> arr=new ArrayList<Integer>();
        boolean prime[] = new boolean[n+1]; 
        for(int i=2;i<n;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=n; p++) 
        { 
            if(prime[p] == true) 
            { 
                arr.add(p);
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
        return arr;
    } 
}