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.BufferedReader;
import java.io.InputStreamReader;
import java.util.Deque;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Scanner;

/**
 * @author avcbcoder last modified @22-Jun-2019 @3:37:24 AM June Circuits 19 -
 *         TODO
 */

public class C {
	public static void main(String[] args) throws Exception {
			solve();
	}

	// **SOLUTION**
	public static void solve() throws Exception {
		int n = sc.nextInt();
		int q = sc.nextInt();

		Deque<Integer> deque = new LinkedList<Integer>();

		int max = -1;
		for (int i = 0; i < n; i++) {
			int x = sc.nextInt();
			deque.addLast(x);
			max = Math.max(max, x);
		}

		HashMap<Integer, String> ans = new HashMap<>();

		int op = 1;
		while (deque.peekFirst() != max) {
			int a = deque.removeFirst();
			int b = deque.removeFirst();
			ans.put(op, a + " " + b);
			deque.addLast(Math.min(a, b));
			deque.addFirst(Math.max(a, b));
			op++;
		}

		deque.removeFirst();
		int[] arr = new int[n - 1];
		for (int i = 0; i < arr.length; i++)
			arr[i] = deque.removeFirst();

		while (q-- > 0) {
			long m = sc.nextLong();
			if (m < op) {
				System.out.println(ans.get((int)m));
			} else {
				long x = m - op;
				x = x % (n - 1);
				System.out.println(max + " " + arr[(int)x]);
			}
		}
	}

	public static InputStreamReader r = new InputStreamReader(System.in);

	public static BufferedReader br = new BufferedReader(r);

	public static Scanner sc = new Scanner(System.in);
}
