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.*;
import java.math.BigInteger;

public class abc{

  public static class FastReader{
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                     InputStreamReader(System.in));
        }

        String n()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String [] nextArray(){
            StringTokenizer tokens = new StringTokenizer(nline(), " ");
            String[] splited = new String[tokens.countTokens()];
            int index = 0;
            while(tokens.hasMoreTokens()){
              splited[index] = tokens.nextToken();
              if(splited[index]!="")
                ++index;
            }
            return splited;
        }

        int ni()
        {
            return Integer.parseInt(n());
        }

        long nl()
        {
            return Long.parseLong(n());
        }

        double nd()
        {
            return Double.parseDouble(n());
        }

        String nline()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
  }


    public static void main (String args[]) throws IOException{
       //Reader in = new Reader();
       FastReader in=new FastReader();
       BufferedOutputStream out=new BufferedOutputStream(System.out);
       StringBuilder sb = new StringBuilder();

       int i,j;
       int n =in.ni(),q=in.ni();
       int a[]=new int[n],b[]=new int[n],c[]=new int[n];
       Deque<Integer> deque = new LinkedList<Integer>();
       for(i=0;i<n;i++){
         a[i]=in.ni();
         c[i]=a[i];
         if(i>0){
           if(a[i]<c[i-1]) c[i]=c[i-1];
         }
       }
       int cur=a[0],ind=0;
       for(i=1;i<n;i++){
         if(cur<a[i]) {
           b[ind++]=cur;
           cur=a[i];
         }
         else{
           b[ind++]=a[i];
         }
       }
       b[n-1]=cur;
       //System.out.println(Arrays.toString(b));
       for(i=0;i<q;i++){
         long m=in.nl();
         if(m<(long)n){
           int k=(int)m;
           sb.append(c[k-1]+" "+a[k]);
         }
         else if(m==(long)n){
           int k=(int)m;
           sb.append(c[k-1]+" "+b[0]);
         }
         else {
           long k=m%(n-1);
           int z=(int)k;
           if(z==0) sb.append(b[n-1]+" "+b[n-2]);
           else sb.append(b[n-1]+" "+b[z-1]);
         }
         sb.append("\n");
       }
       out.write(sb.toString().getBytes());
       out.flush();
    }

    static int[] sieveOfEratosthenes(int n,boolean prime[]) {

        prime[0]=prime[1]=true;
        ArrayList<Integer>list=new ArrayList<Integer>();
        for(int p = 2; p*p <=n; p++){
            // If prime[p] is not changed, then it is a prime
            if(!prime[p] ) {
                // Update all multiples of p
                for(int i = p*p; i <= n; i += p)
                    prime[i] = true;

            }
        }
        list.add(0);
        for(int i = 2; i <= n; i++)
    		{
    			if(!prime[i])
    				list.add(i);
    		}
        return list.stream().mapToInt(x->x).toArray();
    }
    static int gd(int n){
      for(int i=n/2;i>=1;i--) if(n%i==0) return i;
      return -1;
    }

}
