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.*;

public class Main {
  public static Scanner in;

  public static void solve(int n,int q){
    Deque<Integer> deq = new LinkedList<>();
    int max=0,maxIndex=0;    
    
    for(int i=0; i<n; i++){
      deq.add(in.nextInt());
      if(deq.peekLast()>max){
        max=deq.peekLast();
        maxIndex=i;
      }
    }

    int[][] arr = new int[maxIndex][2];

    for(int i=0; i<maxIndex; i++){
      arr[i][0]=deq.pollFirst();
      arr[i][1]=deq.pollFirst();
      
      if(arr[i][0]>arr[i][1]){
        deq.addFirst(arr[i][0]);
        deq.addLast(arr[i][1]);
      }else{
        deq.addFirst(arr[i][1]);
        deq.addLast(arr[i][0]);
      }
    }

    //at this moment max should be in the begining of deq and we can transfer it into arr
    int[] deqArr=new int[n];
    for( int i=0; i<n; i++)    {
      deqArr[i] = deq.pollFirst();
    }
    long[] qr = new long[q];

    for(int i=0; i<q; i++){
        qr[i]=in.nextLong();
    }

    for(int i=0; i<q; i++){
      if(qr[i]-1<maxIndex){
        System.out.printf("%d %d\n",arr[(int)qr[i]-1][0],arr[(int)qr[i]-1][1]);
      }else{
        System.out.printf("%d %d\n",max,deqArr[(int)((qr[i]-1-maxIndex)%(n-1))+1]);

      }
    }
    



  } 

  public static void main(String[] args) {
    

    in = new Scanner(System.in);
    int n = in.nextInt();    
    int q = in.nextInt();    
    solve(n,q);
    
  }
  //while solving this question, before looking at example test cases, I imagined possible consequences and got afraid of it, and therefore didn't solve. 
  //i.e. I gave it up by first impression, without deeper look at it.

}