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

public class C {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        int n = scan.nextInt();
        int q = scan.nextInt();

        LinkedList<Integer> list = new LinkedList<>();
        int max = 0;
        int maxI = 0;

        for (int i = 0; i < n; i++) {
            int next = scan.nextInt();
            list.add(next);

            if (next > max) {
                max = next;
                maxI = i + 1;
            }
        }

        int[][] res = new int[maxI][2];

        for (int i = 0; i < maxI - 1; i++) {
            res[i][0] = list.get(0);
            res[i][1] = list.get(1);

            if (res[i][0] > res[i][1]) {
                list.remove(1);
                list.add(res[i][1]);
            } else {
                list.remove(0);
                list.add(res[i][0]);
            }
        }

        ArrayList<Integer> arrList = new ArrayList<>(list);

        for (int i = 0; i < q; i++) {
            long m = scan.nextLong();

            if (m < maxI) {
                System.out.println(res[(int) (m - 1)][0] + " " + res[(int) (m - 1)][1]);
            } else {
                m -= maxI;
                int i2 = 1 + ((int) (m % (n - 1)));
                System.out.println(arrList.get(0) + " " + arrList.get(i2));
            }
        }
    }
}
