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.io.*;
public class R569D2C {

	public static void main(String[] args) throws Exception{
		// TODO Auto-generated method stub
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
 		
//		int n = Integer.parseInt(bf.readLine());
		StringTokenizer st = new StringTokenizer(bf.readLine());
		int n = Integer.parseInt(st.nextToken());
		int q = Integer.parseInt(st.nextToken());
		int[] deque = new int[n];
		StringTokenizer st1 = new StringTokenizer(bf.readLine());
		int max = 0;
		int maxindex = 0;
		for(int i = 0;i<n;i++){
			deque[i] = Integer.parseInt(st1.nextToken());
			if (deque[i] > max){
				max = deque[i];
				maxindex = i;
			}
		}
		
		ArrayList<Integer> cycleafterMax = new ArrayList<Integer>();
		for(int j = maxindex+1;j<n;j++){
			cycleafterMax.add(deque[j]);
		}
		
		int[][] firstfew = new int[2][maxindex+1];
		int pointer1 = 0;
		int pointer2 = 1;
		for(int i = 0;i<maxindex;i++){
			firstfew[0][i+1] = deque[pointer1];
			firstfew[1][i+1] = deque[pointer2];
			
			if (deque[pointer1] > deque[pointer2]){
				cycleafterMax.add(deque[pointer2]);
				pointer2++;
				
			}
			else{
				cycleafterMax.add(deque[pointer1]);
				
				pointer2++;
				pointer1= pointer2-1;
			}
		}
		
		long[] queries = new long[q];
		for(int i = 0;i<q;i++){
			queries[i] = Long.parseLong(bf.readLine());
		}
		
		int[][] answers = new int[2][q];
		
		for(int i = 0;i<q;i++){
			if (queries[i] <= maxindex){
				answers[0][i] = firstfew[0][(int)(queries[i])];
				answers[1][i] = firstfew[1][(int)(queries[i])];
			}
			else{
				long l = (queries[i] - maxindex-1)%(n-1);
				answers[0][i] = deque[maxindex];
				answers[1][i] = cycleafterMax.get((int)(l));
			}
		}
		for(int i = 0;i<q;i++){
			System.out.print(answers[0][i] + " " + answers[1][i]);
			System.out.println("");
		}
 		
 		
	}

}
