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. 

/**~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
* @author Nicolas Penagos Montoya
* nicolas.penagosm98@gmail.com
**~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
*/

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Deque;
import java.util.LinkedList;

public class Main {

	public static void main(String args[]) throws IOException {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String line = br.readLine();

		String[] parts = line.split(" ");
		Deque<Integer> deque = new LinkedList<Integer>();

		int n = Integer.parseInt(parts[0]);
		int q = Integer.parseInt(parts[1]);
	

		line = br.readLine();
		parts = line.split(" ");

		for (int i = 0; i < n; i++) {
			int e = Integer.parseInt(parts[i]);
			deque.offerLast(e);

		}

		Pair[] uniques = new Pair[n];
		Pair[] loop = new Pair[n-1];
		

		for (int i = 0; i < n; i++) {

			int first = deque.pollFirst();
			int second = deque.pollFirst();
	
			//System.out.println(first+" "+second);
			uniques[i] = new Pair(first, second);

			if (first > second) {
				deque.offerFirst(first);
				deque.offerLast(second);
			} else {
				deque.offerFirst(second);
				deque.offerLast(first);
			}

		}
		//System.out.println();
		for (int i = 0; i < n-1; i++) {
			
			int first = deque.pollFirst();
			int second = deque.pollFirst();
			//System.out.println(first+" "+second);
			loop[i] = new Pair(first, second);

			if (first > second) {
				deque.offerFirst(first);
				deque.offerLast(second);
			} else {
				deque.offerFirst(second);
				deque.offerLast(first);
			}
			
		}
		/*
		System.out.println();
		System.out.println();
		System.out.println();
		
	
		
		
		System.out.println();*/
		while (q > 0) {

			long m = Long.parseLong(br.readLine())-1;

			Pair pair;
			
			if(m<n) {
				pair = uniques[(int)m];
			}else {
				m = m - n;
	
				
				int i = (int)(m%(n-1));
				
		
				pair = loop[i];
			}

			System.out.println(pair.getFirst() + " " + pair.getSecond());

			q--;
		}

	}

}

class Pair {

	private int first;
	private int second;

	public Pair(int f, int s) {
		first = f;
		second = s;
	}

	public int getFirst() {
		return first;
	}

	public void setFirst(int first) {
		this.first = first;
	}

	public int getSecond() {
		return second;
	}

	public void setSecond(int second) {
		this.second = second;
	}

}
