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.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class C {

    void solve(){
       int n = readInt();
       int q = readInt();
       int max = 0;
       int[] a = new int[n];
       Deque<Integer> deque = new ArrayDeque<>();
       for(int i = 0;i<n;i++){
           a[i] = readInt();
           deque.addLast(a[i]);
           max = Math.max(max, a[i]);
       }
        List<String> ans = new ArrayList<>();
        while(deque.peekFirst() != max){
           int one = deque.pollFirst();
           int two = deque.pollFirst();
           ans.add(one + " " + two);
           deque.addFirst(one > two ? one : two);
           deque.addLast(one > two ? two : one);
           if(one == max) break;
       }

       for(int i = 0;i<n;i++){
           a[i] = deque.pollFirst();
       }

       for(int i = 0;i<q;i++){
          long x = readLong();
           if(x <= ans.size()){
               out.println(ans.get((int)x - 1));
               continue;
           }
           x -= ans.size();
           int y =(int) (x%(n - 1) - 1%(n - 1) + (n - 1)) % (n - 1) + 1;
           out.println(max + " " + a[y]);
       }
    }

    public static void main(String[] args) {
        new C().run();
    }

    void run(){
        init();
        solve();
        out.close();
    }

    BufferedReader in;
    PrintWriter out;

    StringTokenizer tok = new StringTokenizer("");

    void init(){
        in = new BufferedReader(new InputStreamReader(System.in));
        out  = new PrintWriter(System.out);
    }

    String readLine(){
        try{
            return in.readLine();
        }catch(Exception ex){
            throw new RuntimeException(ex);
        }
    }
    String readString(){
        while(!tok.hasMoreTokens()){
            String nextLine = readLine();
            if(nextLine == null) return null;
            tok = new StringTokenizer(nextLine);
        }
        return tok.nextToken();
    }

    int readInt(){
        return Integer.parseInt(readString());
    }

    long readLong(){
        return Long.parseLong(readString());
    }

    double readDouble(){
        return Double.parseDouble(readString());
    }
}
