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.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.Deque;

// atharva washimkar
// Jul 04, 2019

public class CODEFORCES_1180_C {

	public static void main (String[] t) throws IOException {
		INPUT in = new INPUT (System.in);
		PrintWriter out = new PrintWriter (System.out);

		int N = in.iscan (), Q = in.iscan ();
		int[] arr = new int[N];

		for (int n = 0; n < N; ++n)
			arr[n] = in.iscan ();

		int max = 0;

		for (int n = 1; n < N; ++n)
			if (arr[n] > arr[max])
				max = n;

		// by the start of the ($max + 1)^th move, A will always be the max element
		// simulate first $max moves
		Deque<Integer> dq = new ArrayDeque<Integer> ();

		for (int n = 0; n < N; ++n)
			dq.offerLast (arr[n]);

		int[] a = new int[max];
		int[] b = new int[max];

		for (int i = 0; i < a.length; ++i) {
			a[i] = dq.peekFirst ();
			// remove a
			int atmp = dq.pollFirst ();
			b[i] = dq.peekFirst ();

			if (a[i] > b[i]) {
				// remove b
				int btmp = dq.pollFirst ();
				dq.offerFirst (atmp);
				dq.offerLast (btmp);
			}
			else {
				dq.offerLast (atmp);
			}
		}

		// after A is the max element, the other elements just keep rotating in order
		int[] cycle = new int[N - 1];
		dq.pollFirst ();

		for (int n = 0; n < cycle.length; ++n)
			cycle[n] = dq.pollFirst ();

		for (int q = 0; q < Q; ++q) {
			long m = in.lscan ();

			if (m <= a.length)
				out.println (a[(int) m - 1] + " " + b[(int) m - 1]);
			else {
				m -= (max + 1);
				m %= (N - 1);
				out.println (arr[max] + " " + cycle[(int) m]);
			}
		}

		out.close ();
	}

	private static class INPUT {

		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar, numChars;

		public INPUT (InputStream stream) {
			this.stream = stream;
		}

		public INPUT (String file) throws IOException {
			this.stream = new FileInputStream (file);
		}

		public int cscan () throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read (buf);
			}

			if (numChars == -1)
				return numChars;

			return buf[curChar++];
		}

		public int iscan () throws IOException {
			int c = cscan (), sgn = 1;

			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			int res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public String sscan () throws IOException {
			int c = cscan ();

			while (space (c))
				c = cscan ();

			StringBuilder res = new StringBuilder ();

			do {
				res.appendCodePoint (c);
				c = cscan ();
			}
			while (!space (c));

			return res.toString ();
		}

		public double dscan () throws IOException {
			int c = cscan (), sgn = 1;

			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			double res = 0;

			while (!space (c) && c != '.') {
				if (c == 'e' || c == 'E')
					return res * UTILITIES.fast_pow (10, iscan ());

				res *= 10;
				res += c - '0';
				c = cscan ();
			}

			if (c == '.') {
				c = cscan ();
				double m = 1;

				while (!space (c)) {
					if (c == 'e' || c == 'E')
						return res * UTILITIES.fast_pow (10, iscan ());

					m /= 10;
					res += (c - '0') * m;
					c = cscan ();
				}
			}

			return res * sgn;
		}

		public long lscan () throws IOException {
			int c = cscan (), sgn = 1;

			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			long res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public boolean space (int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
	}

	public static class UTILITIES {

		static final double EPS = 10e-6;

		public static int lower_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] >= x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int upper_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] > x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int gcd (int a, int b) {
			return b == 0 ? a : gcd (b, a % b);
		}

		public static int lcm (int a, int b) {
			return a * b / gcd (a, b);
		}

		public static int fast_pow_mod (int b, int x, int mod) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;

			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
		}

		public static int fast_pow (int b, int x) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow (b * b, x / 2);

			return b * fast_pow (b * b, x / 2);
		}

		public static long choose (long n, long k) {
			k = Math.min (k, n - k);
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = val * (n - i) / (i + 1);

			return val;
		}

		public static long permute (int n, int k) {
			if (n < k) return 0;
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = (val * (n - i));

			return val;
		}
	}
}