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. 
/* package codechef; // don't place package name! */

import java.io.*;
import java.util.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Codeforces {

    static long[][] m;
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        int n = in.nextInt();
        int q = in.nextInt();
        int max = -1;
        int pos = -1;
        List<Integer> nums = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            int u = in.nextInt();
            nums.add(u);
            if(u > max) {
                max = u;
                pos = i;
            }
        }
        m = new long[pos+1][2];
        for (int i = 0; i < m.length; i++) {
            Arrays.fill(m[i], -1);
        }
        Deque<Integer> dq = new ArrayDeque<>(nums);
        long c = 0;
        int a = -1;
        int b = -1;
        while(true) {
            if(dq.peekFirst() == max) {
                a = dq.pollFirst();
                b = dq.pollFirst();
                m[(int) c][0] = a;
                m[(int) c][1] = b;
                dq.addFirst(a);
                dq.addLast(b);
                break;
            }else {
                a = dq.pollFirst();
                b = dq.pollFirst();
                m[(int) c][0] = a;
                m[(int) c][1] = b;
                if(a > b) {
                    dq.addFirst(a);
                    dq.addLast(b);
                }else {
                    dq.addFirst(b);
                    dq.addLast(a);
                }
                c++;
            }
        }
        c++;
        nums = new ArrayList<Integer>(dq);
        for (int i = 0; i < q; i++) {
            long m = in.nextLong()-1;
            if(m < Codeforces.m.length) {
                out.println(Codeforces.m[(int) m][0] + " " + Codeforces.m[(int) m][1]);
            }else {
                long auxC = c;
                long p = ((m - c) / (n-1)) * (n-1);
                long w = ((m - c) % (n-1));
                int finPos = (int) ((w % (n-1)) + 1);

                out.println(nums.get(0) + " " + nums.get(finPos));
            }
        }
        out.close();
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }
    }
}