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 codeforces;
import java.util.*;
import java.math.*;
public class code_569_C {

	public static class node
	{
		public int data;
		public node link;
		public node(int a,node b)
		{
			data=a;
			link=b;
		}
	}
	static node res ;
	static node rear;
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int q=sc.nextInt();
		int[] arr=new int[n];
		int max=0;
		for(int i=0;i<n;i++)
		{
			arr[i]=sc.nextInt();
			
			if(arr[max]<arr[i])
			{
				max=i;
			}
		}
		
		for(int i=0;i<n;i++)
		{
			if(res==null)
			{
				node ptr=new node(arr[i],null);
				res=ptr;
				rear=res;
			}
			else
			{
				node ptr=new node(arr[i],null);
				rear.link=ptr;
				rear=ptr;
			}
		}
		int[][] cpy=new int[max][2];
		for(int i=0;i<max;i++)
		{
			if(res.data<res.link.data)
			{
				cpy[i][0]=res.data;
				cpy[i][1]=res.link.data;
				node ptr=new node(res.data,null);
				rear.link=ptr;
				rear=ptr;
				
				res=res.link;
				
				
			}
			else
			{
				cpy[i][0]=res.data;
				cpy[i][1]=res.link.data;
				node ptr=new node(res.link.data,null);
				rear.link=ptr;
				rear=ptr;
				res.link=res.link.link;
				
				
			}
		}
		node ptr=res;
		for(int i=0;i<n;i++)
		{
			arr[i]=res.data;
			res=res.link;
		}
		for(int i=0;i<q;i++)
		{
			String s=sc.next();
			if(s.length()<=5)
			{
				int a=Integer.parseInt(s);
				if(a<=max)
				{
					System.out.println(cpy[a-1][0] + " " + cpy[a-1][1]);
				}
				else
				{
					a-=max;
					a%=(n-1);
					if(a==0)
					{
						System.out.println(arr[0] + " " + arr[n-1]);
					}
					else
					{
						System.out.println(arr[0] + " " + arr[a]);
					}
				}
			}
			else
			{
				
				BigInteger b = new BigInteger(s);
				b=b.subtract(new BigInteger(Integer.toString(max)));
				b=b.mod(new BigInteger(Integer.toString(n-1)));
				String str=""+b;
				 
				int a=Integer.parseInt(str);
				if(a==0)
				{
					System.out.println(arr[0] + " " + arr[n-1]);
				}
				else
				{
					System.out.println(arr[0] + " " + arr[a]);
				}
				//System.out.println(a);
			}
			
		}

	}

}
