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.util.*; 
import java.lang.*;
import java.io.*;
public class Main 
{ 
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            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; 
        } 
    } 
    static FastReader s = new FastReader();
    public static void solve(){
        int n = s.nextInt();
        int q = s.nextInt();
        int[] a = new int[n];
        HashMap<Integer,Pair> hashMap = new HashMap<>();
        Deque<Integer> dequeue = new LinkedList<>();
        int maxValue = 0;
        for(int i=0;i<n;i++)
        {
            int temp = s.nextInt();
            dequeue.add(temp);
            maxValue = Math.max(maxValue,temp);
        }
        // System.out.println("maxval "+maxValue);
        int index = 0;
        while(true){
            int temp = dequeue.pollFirst();
            int temp2 = dequeue.pollFirst();
            if(temp==maxValue){
                dequeue.addFirst(temp2);
                dequeue.addFirst(temp);
                // System.out.println("maxCal "+temp);
                break;
            }
            index++;
            hashMap.put(index,new Pair(temp,temp2));
            // System.out.println(hashMap);
            if(temp2>temp){
                dequeue.addFirst(temp2);
                dequeue.addLast(temp);
            }
            else{
                dequeue.addFirst(temp);
                dequeue.addLast(temp2);
            }
        }
        int gar = dequeue.pollFirst();
        for(int i=0;i<n-1;i++){
            a[i] = dequeue.pollFirst();
        }
        for(int i=0;i<q;i++){
            long query = s.nextLong();
            // System.out.println("q "+query);
            if(query<=index){
                System.out.println(hashMap.get((int)query).first+" "+hashMap.get((int)query).second);
            }
            else{
                // System.out.println("i "+(query-index-1)%(n-1));
                System.out.println(maxValue+" "+a[(int)((query-index-1)%(n-1))]);
            }
        }
    }
    public static void main(String[] args) 
    {  
        int T=1;
        // T = s.nextInt();
        while(T>0){
            // System.out.println("TEST CASE: "+T);
            solve();
            T--;
        }
    } 
}
class Pair{
    int first;
    int second;
    Pair(int a,int b){
        this.first = a;
        this.second = b;
    }
    
}
