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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class C {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		
		StringTokenizer st = new StringTokenizer(br.readLine());
		
		int n = Integer.parseInt(st.nextToken()), q = Integer.parseInt(st.nextToken());
		
		st = new StringTokenizer(br.readLine());
		
		LinkedList<Integer> dq = new LinkedList<Integer>();
		
		int max = Integer.MIN_VALUE;
		while(n-->0)
		{
			int a = Integer.parseInt(st.nextToken());
			max = Math.max(max, a);
			dq.add(a);
		}
		
		HashMap<Integer, String> map = new HashMap<Integer, String>();
		
		int[] lastConfig = new int[dq.size()];
		int lastOp = 0;
		for(int i = 0;true;i++)
		{
			int a = dq.removeFirst();
			int b = dq.removeFirst();
			
			map.put(i, a+" "+b);
			
			if(a>b)
			{
				dq.addFirst(a);
				dq.addLast(b);
			}
			else
			{
				dq.addFirst(b);
				dq.addLast(a);
			}
			
			if(a==max||b==max)
			{
				int j = 0;
				for(int num:dq)
					lastConfig[j++] = num;
				lastOp = i;
				break;
			}
		}
		
		
		
		while(q-->0)
		{
			long m = Long.parseLong(br.readLine())-1;
			
			if(m<=Integer.MAX_VALUE&&map.containsKey((int)m))
				pw.println(map.get((int)m));
			else
			{
				long idx = m-lastOp-1;
				
				idx %= (dq.size()-1);
				
//				System.out.println(idx+" "+m+" "+lastOp);
				pw.println(max+" "+lastConfig[(int)(idx+1)]);
			}
				
		}
		
		pw.flush();
		pw.close();
	}
}
