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 round_569;

import java.util.ArrayList;
import java.util.Scanner;

public class deque {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		int q=in.nextInt();
		int[] arr=new int[n];
		int maxi=-1;
		int max=Integer.MIN_VALUE;
		queue que=new queue();
		for(int i=0;i<n;i++) {
			arr[i]=in.nextInt();
			que.enqueue(arr[i]);
			if(arr[i]>max) {
				max=arr[i];
				maxi=i;
			}
		}
		ArrayList<Long> one=new ArrayList<Long>();
		ArrayList<Long> two=new ArrayList<Long>();
		one.add(que.head.val);
		two.add(que.head.next.val);
		for(int j=0;j<maxi;j++) {
			long A=que.head.val;
			long B=que.head.next.val;
			node a=que.head;
			node b=que.head.next;
			if(A<B) {
				que.rear.next=a;
				que.rear=a;
				que.rear.next=null;
				que.head=b;
			}
			else {
				que.rear.next=b;
				que.rear=b;
				que.head=a;
				que.head.next=b.next;
				que.rear.next=null;
			}
			one.add(que.head.val);
			two.add(que.head.next.val);
		}
		ArrayList<Long> abc=new ArrayList<Long>();
		for(int i=0;i<n-1;i++) {
			abc.add(que.head.next.val);
			que.head=que.head.next;
		}
		// System.out.println(abc);
//		System.out.println(que.rear.val);
		for(int i=0;i<q;i++) {
			long m=in.nextLong();
			if(m<=maxi) {					
				System.out.println(one.get((int) (m-1))+" "+two.get((int) (m-1)));
			}
			else {
//				System.out.println(max+" "+abc.get((int) ((m)%(maxi+1))));
				System.out.println(max+" "+abc.get((int) ((m-maxi-1)%abc.size())));
			}
		}
	}

}
class node{
	long val;	
	node next;
	node(){
		val=0;
		next=null;
	}
	node(int x){
		val=x;
		next=null;
	}
}

class queue{
	node head;
	node rear;
	void enqueue(int x){
		if(head==null) {
			head=new node(x);
			rear=head;
		}
		else if(head==rear) {
			rear=new node(x);
			head.next=rear;
		}
		else {
			rear.next=new node(x);
			rear=rear.next;
		}
	}
	long deq() {
		if(head==null) {
			return -1;
		}
		else 
			return head.val;
	}
}