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.util.ArrayDeque;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int queries = in.nextInt();

        ArrayDeque<Integer> deq = new ArrayDeque<Integer>();
        for (int i = 0; i < n; i += 1) {
            deq.addLast(in.nextInt());
        }

        Solver solver = new Solver(deq);
        for (int i = 0; i < queries; i += 1) {
            System.out.println(solver.query(in.nextLong()));
        }
    }
}

class Solver {
    private Pair[] smallRes;
    private Integer[] finalArray;

    public Solver(ArrayDeque<Integer> deq) {
        int n = deq.size();

        smallRes = new Pair[n - 1];
        for (int i = 0; i + 1 < n; i += 1) {
            int a = deq.pollFirst();
            int b = deq.pollFirst();

            smallRes[i] = new Pair(a, b);
            deq.addFirst(Math.max(a, b));
            deq.addLast(Math.min(a, b));
        }

        finalArray = new Integer[deq.size()];
        deq.toArray(finalArray);
    }

    public Pair query(long turn) {
        int n = finalArray.length;
        if (turn < n) {
            return new Pair(smallRes[(int)turn - 1]);
        }

        int first = finalArray[0];
        int second = finalArray[1 + (int)((turn - n) % (n - 1))];

        return new Pair(first, second);
    }
}

class Pair {
    public int first;
    public int second;

    public Pair() {
        this(0, 0);
    }

    public Pair(int a, int b) {
        first = a;
        second = b;
    }

    public Pair(Pair other) {
        first = other.first;
        second = other.second;
    }

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