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.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.LinkedList;
import java.util.List;
import java.util.StringTokenizer;

public class ValeriyAndDeque { // Template for CF
    public static class ListComparator implements Comparator<List<Integer>> {

        @Override
        public int compare(List<Integer> l1, List<Integer> l2) {
            for (int i = 0; i < l1.size(); ++i) {
                if (l1.get(i).compareTo(l2.get(i)) != 0) {
                    return l1.get(i).compareTo(l2.get(i));
                }
            }
            return 0;
        }
    }

    public static class Pair {
        int first;
        int second;

        public Pair(int first, int second) {
            this.first = first;
            this.second = second;
        }

        public int getFirst() {
            return first;
        }

        public int getSecond() {
            return second;
        }

        @Override
        public String toString() {
            return first + " " + second;
        }
    }

    public static void main(String[] args) throws IOException {
        // Check for int overflow!!!!
        // Should you use a long to store the sum or smthn?
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        StringTokenizer st = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(st.nextToken());
        int q = Integer.parseInt(st.nextToken());
        int max = -1;
        LinkedList<Integer> queue = new LinkedList<>();
        st = new StringTokenizer(f.readLine());
        for (int i = 0; i < n; i++) {
            int a = Integer.parseInt(st.nextToken());
            max = Math.max(a, max);
            queue.add(a);
        }
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < q; i++) {
            long a = Long.parseLong(f.readLine());
            list.add(a);
        }
        long count = 0;
        List<Pair> pairs = new ArrayList<>();
        while (true) {
            int a = queue.pollFirst();
            int b = queue.pollFirst();
            count++;
            pairs.add(new Pair(a, b));
            if (a > b) {
                queue.addFirst(a);
                queue.addLast(b);
            } else {
                queue.addFirst(b);
                queue.addLast(a);
            }
            if (a == max || b == max) {
                break;
            }

        }

        List<Integer> fList = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            fList.add(queue.pollFirst());
        }

        // System.out.println(fList);
        // System.out.println(pairs);
        for (int i = 0; i < q; i++) {
            if (list.get(i) <= count) {
                out.println(pairs.get((int) (list.get(i) - 1)).getFirst() + " "
                        + pairs.get((int) (list.get(i) - 1)).getSecond());
            } else {
                long a = (list.get(i) - count) % (n - 1);
                // System.out.println(a);
                if (a == 0) {
                    a = n - 1;
                }
                out.println(fList.get(0) + " " + fList.get((int) a));
            }
        }
        out.close();
    }

}