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.util.ArrayList;
import java.util.StringTokenizer;
public class Tram {
	public static void main(String[] args) {
		try {
			BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
			StringTokenizer st = new StringTokenizer(br.readLine());
		    int n = Integer.parseInt(st.nextToken());
		    int q= Integer.parseInt(st.nextToken());
		    if(q>0) {
		    	ArrayList<Long> deque = new ArrayList<Long>();
		    	st = new StringTokenizer(br.readLine());
		    	long max = 0;
		    	for(int i=0;i<n;i++) {
		    		Long x = Long.parseLong(st.nextToken());
		    		deque.add(x);
		    		if(x>max) {
		    			max =x;
		    		}
		    	}
		    	ArrayList<long[]> pareja = new ArrayList<long[]>();
		    	long a,b;
		    	while(deque.get(0)<max) {
		    		if((a=deque.get(0).longValue())< (b=deque.get(1).longValue())){
		    			deque.add(deque.remove(0));
		    		}else {
		    			deque.add(deque.remove(1));
		    		}
		    		long p[]  = {a,b};
		    		pareja.add(p);
		    	}
		    	for(int i=0; i<q;i++) {
		    		long pos = Long.parseLong(br.readLine());
		    		if(pos<= pareja.size()) {
		    			long[] p = pareja.get((int) pos-1);
		    			System.out.println(p[0]+" "+p[1]);
		    		}else {
		    			int pos2 = (int)(pos%(deque.size()-1))-pareja.size();
		    			if(pos2<=0) {
		    				pos2+=deque.size()-1;
		    				if(pos2==0) {
		    					pos2=deque.size()-1;
		    				}
		    			}
		    			System.out.println(deque.get(0)+" "+deque.get(pos2));
		    		}
		    	}
		    }
		    br.close();
		} catch (IOException e) {
			e.printStackTrace();
		}
	}
}
