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.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.StringTokenizer;

public class cf2 {
	static final double EPS = 1e-9;

	static int[] isComposite;
	static ArrayList<Integer> primes;
	static int n;
	static int k;
	static int y[];

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

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int x[] = new int[n];
		int max = -1;
		for (int i = 0; i < n; i++) {
			x[i] = sc.nextInt();
			max = Math.max(max, x[i]);
		}

		int y[] = new int[n-1];
		int left[] = new int[n + 1];
		int right[] = new int[n + 1];
		for (int i = 0; i < n - 1; i++) {
			left[i + 1] = x[i];
			right[i + 1] = x[i + 1];
			if (x[i] > x[i + 1]) {
				y[i] = x[i + 1];
				x[i + 1] = x[i];
			} else {
				y[i] = x[i];
			}
		}

//		System.out.println(Arrays.toString(y));
		StringBuilder sb = new StringBuilder();
		while (m-- > 0) {
           long q = sc.nextlong();
			if(q<n) {
				sb.append(left[(int)q]+ " "+right[(int)q]+"\n");
			}else {
				q-=n;
				sb.append(max+" "+y[(int)(((q)%(n-1)))]+"\n");
			}
//				1 2 3 4 
		}
		System.out.println(sb.toString());

	}

	public static void printAns() {

	}

	public static class Point implements Comparable<Point> {

		static final double EPS = 1e-9;

		int x, y;
		// String c;

		Point(int a, int b) {
			x = a;
			y = b;
			// c = s;
		}

		public int compareTo(Point p) {
			if (Math.abs(x - p.x) > EPS)
				return x > p.x ? 1 : -1;
			if (Math.abs(y - p.y) > EPS)
				return y > p.y ? 1 : -1;
			return 0;
		}

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

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

		public Scanner(String file) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(file));
		}

		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public long nextlong() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			long res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}
}
