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 static java.lang.Math.*;
import static java.util.Arrays.*;

import java.util.*;
import java.io.*;

public class A {

    void main() throws Exception
    {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out) ;
        int n = sc.nextInt() ;
        int TC = sc.nextInt() ;
        int [] a = new int[n] ;
        LinkedList<Integer> q = new LinkedList<>() ;
        boolean same = true ;
        int query [][] = new int [n << 4 | 1][2] ;
        int max = 0 ;
        for(int i = 0 ; i < n ;i++) {
            a[i] = sc.nextInt();
            max = max(max , a[i]) ;
            if(i > 0) same &= a[i] == a[i - 1] ;
        }
        if(same)
        {
            while(TC-->0)
            {
                sc.nextLong() ;
                out.println(a[0]+" "+a[0]);
            }
        }
        else
        {
            for(int i = 0 ; i < n ;i++)
                q.add(a[i]) ;
            int opp = 1 ;
            while(true)
            {

                int A = q.pollFirst() ;
                int B = q.pollFirst();
                if(A > B && A == max)
                {
                    q.addFirst(B);
                    q.addFirst(A);
                    break;
                }
                query[opp ++] = new int []{A , B} ;

                if(A > B)
                {
                    q.addFirst(A);
                    q.addLast(B);
                }
                else
                {
                    q.addFirst(B);
                    q.addLast(A);
                }

            }
            int idx = 0 ;
            while(!q.isEmpty())
            {
                int x = q.pollFirst() ;
                a[idx ++] = x ;
            }
            int size = n - 1 ;

            while(TC-->0)
            {
                long t = sc.nextLong() ;
                if(t < opp)
                    out.println(query[(int)(t)][0] + " "+query[(int)(t )][1]);
                else
                {
                    int to = (int)((t - opp) % size) + 1 ;
                    out.println(a[0]+" "+a[to]);
                }
            }
        }

        out.flush();
        out.close();
    }

    class Scanner
    {
        BufferedReader br;
        StringTokenizer st;

        Scanner(InputStream in)
        {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        Scanner(String path) throws Exception{
            br = new BufferedReader((new FileReader(path))) ;
        }
        String next() throws Exception
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        int nextInt() throws Exception { return Integer.parseInt(next()); }

        long nextLong() throws Exception { return Long.parseLong(next()); }

        double nextDouble() throws Exception { return Double.parseDouble(next());}
    }

    public static void main (String [] args) throws Exception {(new A()).main();}

}