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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.Deque;
import java.util.LinkedList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author real
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            Deque<Integer> qu = new LinkedList<>();
            int ar[] = in.nextIntArray(n);
            for (int i = 0; i < n; i++) {
                qu.addLast(ar[i]);
            }
            int ans[][] = new int[n - 1][2];
            for (int i = 0; i < n - 1; i++) {
                int first = qu.pollFirst();
                int second = qu.pollFirst();
                int mx = Math.max(first, second);
                int mn = Math.min(first, second);
                qu.addFirst(mx);
                qu.addLast(mn);
                ans[i][0] = first;
                ans[i][1] = second;
            }
            int val[] = new int[n - 1];
            int first = qu.pollFirst();
            for (int i = 0; i < n - 1; i++) {
                val[i] = qu.pollFirst();
            }

            for (int i = 0; i < m; i++) {
                long q = in.nextLong();
                if (q >= n) {
                    q = q - n;
                    out.println(first + " " + val[(int) (q % (n - 1))]);
                } else {
                    out.println(ans[(int) (q - 1)][0] + " " + ans[(int) (q - 1)][1]);
                }
            }

        }

    }

    static class InputReader {
        private final InputStream stream;
        private final byte[] buf = new byte[8192];
        private int curChar;
        private int snumChars;

        public InputReader(InputStream st) {
            this.stream = st;
        }

        public int read() {
            //*-*------clare------
//remeber while comparing 2 non primitive data type not to use ==
//remember Arrays.sort for primitive data has worst time case complexity of 0(n^2) bcoz it uses quick sort
//again silly mist[[[[[[pakes ,yr kb tk krta rhega ye mistakes
//try to write simple codes ,break it into simple things
            // for test cases make sure println(); ;)
//knowledge>rating
        /*
        public class Main
        implements Runnable{
    public static void main(String[] args) {
        new Thread(null,new Main(),"Main",1<<26).start();

    }
    public void run() {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();//chenge the name of task
        solver.solve(1, in, out);
        out.close();
    }
         */

            if (snumChars == -1)
                throw new InputMismatchException();
            if (curChar >= snumChars) {
                curChar = 0;
                try {
                    snumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (snumChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        public int nextInt() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public long nextLong() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public int[] nextIntArray(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }

        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

    }
}

