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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;

public class Codeforces {

    static class Pair {
        int a;
        int b;

        public Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }
    }

    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[] arr = in.nextIntArr(n);
        long[] queries = in.nextLongArr(q);

        int maxIndex = 0;
        int maxVal = arr[0];
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] > maxVal) {
                maxVal = arr[i];
                maxIndex = i;
            }
        }

        Deque<Integer> queue = new LinkedList<>();
        for (int i = 0; i < arr.length; i++) {
            queue.add(arr[i]);
        }

        Map<Long, Pair> resMap = new HashMap<>();
        for (int i = 0; i < maxIndex; i++) {
            int a = queue.poll();
            int b = queue.poll();

            if (a > b) {
                queue.addFirst(a);
                queue.addLast(b);
            } else {
                queue.addFirst(b);
                queue.addLast(a);
            }

            resMap.put((long) i + 1, new Pair(a, b));
        }

        int[] newArr = new int[n];
        for (int i = 0; i < newArr.length; i++) {
            newArr[i] = queue.poll();
        }

        for (int i = 0; i < queries.length; i++) {
            long query = queries[i];

            if (query >= maxIndex + 1) {
                int a = newArr[0];
                int b = newArr[(int) ((query - maxIndex - 1) % (n - 1)) + 1];

                out.println(a + " " + b);
            } else {
                Pair pair = resMap.get(query);
                out.println(pair.a + " " + pair.b);
            }
        }

        out.close();
    }

    private static void swap(int[] arr, int i, int j) {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }

    private static void printArr(int[] arr, PrintWriter out) {
        for (int i = 0; i < arr.length; i++) {
            out.print(arr[i] + " ");
        }
        out.println();
    }

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

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            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 String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }

            return null;
        }

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

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

        public int[] nextIntArr(int n) {
            int[] arr = new int[n];
            for (int j = 0; j < arr.length; j++) {
                arr[j] = nextInt();
            }

            return arr;
        }

        public long[] nextLongArr(int n) {
            long[] arr = new long[n];
            for (int j = 0; j < arr.length; j++) {
                arr[j] = nextLong();
            }

            return arr;
        }
    }
}