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.io.*;
import java.util.*;
public class Main
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int queries=sc.nextInt();
        long a[]=new long[n];
        int max_index=0;
        Deque<Long> q=new LinkedList<>();
        for( int i=0;i<n;i++)
        {
            a[i]=sc.nextLong();
            q.add(a[i]);
            if(a[i]>a[max_index])
            max_index=i;
        }
       
        long ans[][]=new long[max_index+1][2];
        for( int i=0;i<=max_index;i++)
        {
            long val1=q.removeFirst();
            //System.out.println(val1);

            long val2=q.removeFirst();
           // System.out.println(val2);
            ans[i][0]=val1;
            ans[i][1]=val2;
            if(val1>val2)
            {
                q.push(val1);
                q.addLast(val2);
            }
            else
            {
                q.push(val2);
                q.addLast(val1);
            }
            
        }
        ArrayList<Long> adj=new ArrayList<>();
        while(!q.isEmpty())
        {
            adj.add(q.pop());
        }
    
        while(queries-->0)
        {
            long m=sc.nextLong();
            if(m<=max_index+1)
            {
                System.out.println(ans[(int)m-1][0]+" "+ans[(int)m-1][1]);
            }
            else
            {
                m-=max_index+1;
                m%=(n-1);
                if(m==0)
                m=n-1;
                System.out.println(adj.get(0)+" "+adj.get((int)m));
            }
        }

    }
}