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 b {
    FastScanner in;
    PrintWriter out;
    boolean systemIO = true;
    LinkedList<Long> ls;
    int n, INF = Integer.MAX_VALUE / 2;

    void pr() {
        LinkedList<Long> l = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            long a = ls.removeFirst();
            System.out.print(a + " ");
            l.addLast(a);
        }
        System.out.println();
        ls = l;
    }

    void solve() {
        n = in.nextInt();
        int k = in.nextInt();
        ls = new LinkedList<>();
        long max = -INF;
        for (int i = 0; i < n; i++) {
            long u = in.nextLong();
            ls.add(u);
            max = Math.max(max, u);
        }
        ArrayList<p> ans1 = new ArrayList<>();
        while (ls.peekFirst() != max) {
            //pr();
            long a = ls.removeFirst();
            long b = ls.removeFirst();
            ans1.add(new p(a, b));
            if (a > b) {
                ls.addFirst(a);
                ls.addLast(b);
            } else {
                ls.addFirst(b);
                ls.addLast(a);
            }
        }
        p ans2[] = new p[n - 1];
        for (int i = 0; i < n - 1; i++) {
            //pr();
            long a = ls.removeFirst();
            long b = ls.removeFirst();
            ans2[i] = (new p(a, b));
            if (a > b) {
                ls.addFirst(a);
                ls.addLast(b);
            } else {
                ls.addFirst(b);
                ls.addLast(a);
            }
        }
        for (int i = 0; i < k; i++) {
            long u = in.nextLong() - 1;
            if (u < ans1.size()) {
                out.println(ans1.get((int) u).a + " " + ans1.get((int) u).b);
            } else {
                u = u - ans1.size() ;
                u %= (n - 1);
                out.println(ans2[(int) u].a + " " + ans2[(int) u].b);
            }
        }


    }

    class p {
        long a, b;

        p(long A, long B) {
            this.a = A;
            this.b = B;
        }
    }

    private void run() throws IOException {
        if (systemIO) {
            in = new b.FastScanner(System.in);
            out = new PrintWriter(System.out);
        } else {
            in = new b.FastScanner(new File("input.txt"));
            out = new PrintWriter(new File("output.txt"));
        }
        solve();

        out.close();
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

        FastScanner(InputStream f) {
            br = new BufferedReader(new InputStreamReader(f));
        }

        String nextLine() {
            try {
                return br.readLine();
            } catch (IOException e) {
                return null;
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

    }

    public static void main(String[] arg) throws IOException {
        new b().run();
    }

}