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.*;
import java.io.*;

public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		
		int n = in.nextInt();
		int q = in.nextInt();
		int[] a = new int[2 * n];
		for(int i = 0; i < n; i++)
		{
			a[i] = in.nextInt();
		}
		
		int max = -1;
		for(int x : a)
			max = Math.max(max, x);
		
		int start = 0;
		int next = 1;
		int[] a1 = new int[n + 1];
		int[] a2 = new int[n + 1];
		while(a[start] != max)
		{
			int x = a[start++];
			int y = a[start];
			a1[next] = x;
			a2[next] = y;
			a[start] = Math.max(x, y);
			a[start + n - 1] = Math.min(x, y);
			next++;
		}
		
		for(int i = 0; i < q; i++)
		{
			long w = in.nextLong();
			if(w < next)
			{
				out.println(a1[(int)w] + " " + a2[(int)w]);
			}
			else
			{
				long s = w - next;
				s %= (n - 1);
				out.println(a[start] + " " + a[(int)(start + 1 + s)]);
			}
		}
		
		
		out.close();
			
	}
	
	static class InputReader
	{
		BufferedReader reader;
		StringTokenizer tokenizer;
		
		public InputReader(InputStream stream)
		{
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}
		
		public String next()
		{
			while(tokenizer == null || !tokenizer.hasMoreTokens())
			{
				try
				{
					tokenizer = new StringTokenizer(reader.readLine());
				}
				catch(IOException e)
				{
					throw new RuntimeException(e);
				}
			}
			
			return tokenizer.nextToken();
		}
		
		public int nextInt()
		{
			return Integer.parseInt(next());
		}
		
		public long nextLong()
		{
			return Long.parseLong(next());
		}
	}
}