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.Arrays;
import java.util.Deque;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class Main1 {

	static FastReader input = new FastReader();
	static PrintWriter out = new PrintWriter(System.out);
	static int m = 0;

	public static void main(String[] args) throws IOException {

		int n = input.nextInt();
		int m = input.nextInt();

		Deque<Integer> dq = new LinkedList<Integer>();

		ArrayList<int[]> bef = new ArrayList<>();
		ArrayList<int[]> aft = new ArrayList<>();

		for (int i = 0; i < n; i++) {
			dq.add(input.nextInt());
		}

		int max = Integer.MIN_VALUE;

		for (Integer x : dq)
			max = Math.max(x, max);


		while (dq.getFirst() != max) {

			int x = dq.poll();
			int y = dq.poll();

			bef.add(new int[] { x, y });

			dq.push(Math.max(x, y));
			dq.add(Math.min(x, y));
		}

		for (int i = 0; i < n - 1; i++) {
			int x = dq.poll();
			int y = dq.poll();

			aft.add(new int[] { x, y });

			dq.push(Math.max(x, y));
			dq.add(Math.min(x, y));
		}

		for (int i = 0; i < m; i++) {
			long x = input.nextLong() - 1;
			int a, b;

			if (x < bef.size()) {
				a = bef.get((int) x)[0];
				b = bef.get((int) x)[1];
			} else {
				x -= bef.size();
				int pos = (int) (x % (n - 1));
				a = aft.get(pos)[0];
				b = aft.get(pos)[1];
			}
			out.println(a + " " + b);
		}
		out.flush();

	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreElements()) {
				st = new StringTokenizer(br.readLine());
			}
			return st.nextToken();
		}

		int nextInt() throws NumberFormatException, IOException {
			return Integer.parseInt(next());
		}

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		String nextLine() throws IOException {
			String str = "";
			str = br.readLine();
			return str;
		}
	}

}