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. 
/* package whatever; // don't place package name! */

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 Ideone
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		long q = s.nextLong();
		long max = -1;
		int max_index = -1;
		Deque<Long> A= new LinkedList<Long>();
		for(int i = 0;i < n;i++){
			long a = s.nextLong();
			A.add(a);
			if(a > max){
				max = a;
				max_index = i;
			}
		}
		
		Map<Integer, List<Long>> result = new HashMap<>();
		int y = max_index + 1;
		for(int i = 1; i < y ;i++){
			   long a = A.poll();
			  long b = A.poll();
			  result.put(i, Arrays.asList(a, b));
			  if(a > b){
			  	A.addFirst(a);
			  	A.add(b);
			  }else{
			  	A.addFirst(b);
			  	A.add(a);
			  
			  }
		}
		long B[] = new long[n];
		for(int i = 0;i < n;i++){
			B[i] = A.poll();
		}
		int k = n - 1;

		while(q > 0){
		   long a;
		   a = s.nextLong();
		  if(a < y){
		  	int x = (int)a;
			System.out.println(result.get(x).get(0) + " " + result.get(x).get(1));
		  }else {
		  	long x = (a - y);
		  	x = x % k ;
		  	x++;
		  	System.out.println(max + " " + B[(int)x]);
		  }
	    q--;
		}
		s.close();
		
	}
}