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.io.*;
import java.util.*;
public class C {
	
	public static void main(String[] args) {
		InputReader in = new InputReader();
		int n = in.nextInt();
		int Q = in.nextInt();
		
		int max = Integer.MIN_VALUE, maxPos = -1;
		int[] arr = new int[n];
		for (int i=0; i<n; i++) {
			arr[i] = in.nextInt();
			if (arr[i] > max) {
				max = arr[i];
				maxPos = i;
			}
		}
		
		int[][] beforeMax = new int[maxPos+1][2];
		ArrayList<Integer> list = new ArrayList<>();
		int pos = 1, a = arr[0], b = arr[1];
		for (int i=1; i<=maxPos; i++) {
			beforeMax[i][0] = a;
			beforeMax[i][1] = b;
			list.add(Math.min(a, b));
			if (i<maxPos) {
				a = Math.max(a, b);
				b = arr[++pos];
			}
		}
		
		int[] newArr = reOrder(arr, list, maxPos);
		
		StringBuffer sb = new StringBuffer();
		for (int i=0; i<Q; i++) {
			long q = in.nextLong();
			if (q <= maxPos)
				sb.append(beforeMax[(int) q][0] + " " + beforeMax[(int) q][1] + "\n");
			else {
				q -= (maxPos+1);
				q %= (n-1);
				sb.append(max + " " + newArr[(int) q] + "\n");
			}
		}
		System.out.println(sb);
	}
	
	public static int[] reOrder(int[] arr, ArrayList<Integer> list, int maxPos) {
		int[] newArr = new int[arr.length-1];
		for (int i=maxPos+1; i<arr.length; i++)
			newArr[i-(maxPos+1)] = arr[i];
		for (int i=0; i<maxPos; i++)
			newArr[i+arr.length-(maxPos+1)] = list.get(i);
		return newArr;
	}

	static class InputReader {
		public BufferedReader br;
		public StringTokenizer st;
		
		public InputReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
			st = null;
		}
		
		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return st.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
		
		public long nextLong() {
			return Long.parseLong(next());
		}
	}
}