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.ArrayDeque;
import java.util.HashMap;
import java.util.Scanner;

public class VALERY_AND_DEUEUE {

	public static void main(String[] args) {

		Scanner nik = new Scanner(System.in);
		int n = nik.nextInt();
		int q = nik.nextInt();

		ArrayDeque<Integer> q1 = new ArrayDeque<>();
		int max = 0;
		int maxi = 0;
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = nik.nextInt();
			q1.addLast(a[i]);
			if (a[i] > max) {
				max = a[i];
				maxi = i;
			}
		}
		// System.out.println(q1);
		// System.out.println(max + " " + maxi);
		HashMap<Long, pair> ht = new HashMap<>();

		for (long i = 0; i < maxi; i++) {
			int temp1 = q1.removeFirst();
			int temp2 = q1.removeFirst();
			if (temp1 > temp2) {
				q1.addFirst(temp1);
				q1.addLast(temp2);
			} else {
				q1.addFirst(temp2);
				q1.addLast(temp1);
			}
			ht.put(i, new pair(temp1, temp2));
		}

		// for (Long val : ht.keySet()) {
		// pair temp = ht.get(val);
		// System.out.println(temp.st + " " + temp.ed);
		// }
		HashMap<Long, pair> hm = new HashMap<>();

		for (long i = 0; i < (n - 1); i++) {
			int temp1 = q1.removeFirst();
			int temp2 = q1.removeFirst();
			if (temp1 > temp2) {
				q1.addFirst(temp1);
				q1.addLast(temp2);
			} else {
				q1.addFirst(temp2);
				q1.addLast(temp1);
			}
			hm.put(i, new pair(temp1, temp2));
		}
		// System.out.println(ht);
		// System.out.println(hm);
		while (q-- > 0) {
			long m = nik.nextLong() - 1;

			if (m < maxi) {
				pair temp = ht.get(m);
				int temp1 = temp.st;
				int temp2 = temp.ed;
				System.out.println(temp.st + " " + temp.ed);
			} else {
				m -= maxi;
				m = m % (n - 1);

				pair temp = hm.get(m);
				System.out.println(temp.st + " " + temp.ed);
			}
		}

	}

	private static class pair {

		int st;
		int ed;

		pair(int st, int ed) {
			this.st = st;
			this.ed = ed;
		}
	}
}
