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.*; 
import java.util.*;
  
public class Main {
    static class FastScanner
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
  
    public static void main(String[] args) throws IOException {
        FastScanner sc=new FastScanner();
        int N = sc.nextInt();
        int Q = sc.nextInt();
        Deque<Integer> d = new LinkedList<Integer>();
        int maxElement = 0;
        int maxI = -1;
        for (int i = 0; i < N; i++) {
        	int n = sc.nextInt();
        	d.add(n);
        	if (n > maxElement) {
        		maxElement = n;
        		maxI = i;
        	}
        }
        if (Q == 0)
        	return;
        long[][] queries = new long[Q][2];
        for (int i = 0; i < Q; i++) {
        	queries[i][0] = sc.nextLong();
        	queries[i][1] = i;
        }
        Arrays.sort(queries, new Comparator<long[]>() {
        	  @Override
        	  public int compare(long[] arr1, long[] arr2) {
        	    if (arr1[0] < arr2[0])
        	    	return -1;
        	    else
        	    	return 1;
        	  }
        });
        int q = 0;
        int[][] ans = new int[Q][2];
        //Simulate first maxI operations of the deque
        for (int i = 1; i <= maxI; i++) {
        	int A = d.pop();
        	int B = d.pop();
        	while ((int)queries[q][0] == i) {
        		ans[(int)queries[q][1]][0] = A;
        		ans[(int)queries[q][1]][1] = B;
        		q++;
        		if (q == Q) break;
        	}
        	if (A > B) {
        		d.addFirst(A);
        		d.addLast(B);
        	} else {
        		d.addFirst(B);
        		d.addLast(A);
        	}
        }
        if (q < Q) {
            long operation = (long)maxI+1;
            int[] rotating = new int[N-1];
            int top = d.pop();
            for (int i = 0; i < N-1; i++)
            	rotating[i] = d.pop();
            while (q < Q) {
            	long jump = queries[q][0] - operation;
        		ans[(int)queries[q][1]][0] = top;
        		long index = jump - (N-1) * (jump/(N-1));
        		ans[(int)queries[q][1]][1] = rotating[(int)index];
        		q++;
            }
        }
        for (int[] a: ans) {
        	System.out.println(a[0] + " " + a[1]);
        }
    }
}