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 //do not write package name here */

import java.io.*;
import java.util.*;
public class S {
	public static void main (String[] args) {
		Scanner in=new Scanner(System.in);
		int n=in.nextInt(),q=in.nextInt();
		int a[]=new int[n];
		int max=Integer.MIN_VALUE;
		Deque<Integer> dq=new LinkedList<>();
		for(int i=0;i<n;i++)
		{
		    a[i]=in.nextInt();
		    dq.add(a[i]);
		    max=Math.max(max,a[i]);
		}   
		ArrayList<point> l1=new ArrayList<>();
		ArrayList<point> l2=new ArrayList<>();
		//l2.add(new point(0,0));
		while(dq.peek()!=max)
		{
		    int r=dq.removeFirst();
		    int s=dq.removeFirst();
		    dq.addFirst(Math.max(r,s));
		    dq.addLast(Math.min(r,s));
		    l1.add(new point(r,s));
		}
		dq.removeFirst();
		while(!dq.isEmpty())
		{
		    l2.add(new point(max,dq.removeFirst()));
		}
		while(q--!=0)
		{
		    long m=in.nextLong();
		    if(m<=l1.size())
		    {
		        point g=l1.get((int)(m-1));
		        System.out.println(g.x+" "+g.y);
		    }      
		    else
		    {
		        m-=(l1.size());
		        m%=(l2.size());
		        if(m==0)
		        {
		            point g=l2.get(l2.size()-1);
		            System.out.println(g.x+" "+g.y);
		            continue;
		        }
		        point g=l2.get((int)(m-1));
		        System.out.println(g.x+" "+g.y);
		    }
		}
	}
}
class point
{
    int x,y;
    public point(int x,int y)
    {
        this.x=x;
        this.y=y;
    }
}