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.io.*;
import java.math.*;

public class Main
{

    public static void process()throws IOException
    {
        int n=ni(),q=ni(),mx_idx=1;
        int arr[]=new int[n+1];
        for(int i=1;i<=n;i++) arr[i]=ni();
        for(int i=1;i<=n;i++)
            if(arr[i]>arr[mx_idx])
                mx_idx=i;

        String ans[]=new String[n+1];

        ArrayList<Integer> li = new ArrayList<>();
        for(int i=mx_idx+1;i<=n;i++) li.add(arr[i]);

        int take=1;
        for(int i=1;i<mx_idx;i++){
            ans[i]=arr[take]+" "+arr[i+1];

            if(arr[take]<arr[i+1]){
                li.add(arr[take]);
                take=i+1;
            }
            else
                li.add(arr[i+1]);
        }

        // if(mx_idx==n)
        //     ans[mx_idx]=arr[mx_idx]+" "+li.get(0);
        // else
        //     ans[mx_idx]=arr[mx_idx]+" "+arr[mx_idx+1];

        //System.out.println(li);System.out.println(Arrays.toString(ans));
        //System.out.println("max ele "+arr[mx_idx]);
        while(q-- > 0){
            long x=nl();
            if(x<mx_idx)
                pn(ans[(int)x]);
            else{
                x=(x-mx_idx)%(n-1);
                pn(arr[mx_idx]+" "+li.get((int)x));
            }
        }        
    }


    static FastReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        out = new PrintWriter(System.out);
        sc=new FastReader();

        long s = System.currentTimeMillis();
        int t=1;
        //t=ni();
        while(t-->0)
            process();

        out.flush();
        System.err.println(System.currentTimeMillis()-s+"ms");
    }
    
    
    static void pn(Object o){out.println(o);}
    static void p(Object o){out.print(o);}
    static int ni()throws IOException{return Integer.parseInt(sc.next());}
    static long nl()throws IOException{return Long.parseLong(sc.next());}
    static double nd()throws IOException{return Double.parseDouble(sc.next());}
    static String nln()throws IOException{return sc.nextLine();}
    static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}
    
    static long mod=(long)1e9+7l;

    static<T> void r_sort(T arr[],int n){
        Random r = new Random(); 
        for (int i = n-1; i > 0; i--){ 
            int j = r.nextInt(i+1); 
                  
            T temp = arr[i]; 
            arr[i] = arr[j]; 
            arr[j] = temp; 
        } 
        Arrays.sort(arr); 
    }
    
/////////////////////////////////////////////////////////////////////////////////////////////////////////
    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(); 
        } 
  
        String nextLine(){ 
            String str = ""; 
            try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } 
            return str; 
        } 
    } 
/////////////////////////////////////////////////////////////////////////////////////////////////////////////
}