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.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;

//import Number_theory.pair;

import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

//import graph_representation.edge;
public class hacker49 {
	public static int r1=0;
	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 	ArrayList<Integer>[] f=new ArrayList[400001];
static int[] vis=new int[400001];
static 	int[] deg=new int[100001];
	static HashMap<Long,Integer> e=new HashMap<>();
	public static void main(String[] args) {
		OutputStream outputStream =System.out;
	    PrintWriter out =new PrintWriter(outputStream);
		FastReader s=new FastReader();
int n=s.nextInt();
int q=s.nextInt();
//long[] a=new long[n+1];
ArrayList<Long> e=new ArrayList<>();
long max=Long.MIN_VALUE;
for(int i=0;i<n;i++) {
//	e.add(s.nextLong());
	long x=s.nextLong();
	e.add(x);
	max=Math.max(x,max);
}
ArrayList<pair> e1=new ArrayList<>();
int count=0;
out:for(int i=0;i<n;i++) {
	if(e.get(0)==max) {
	break out;	
	}else {
		e1.add(new pair(e.get(0),e.get(1)));
		if(e.get(0)>=e.get(1)) {
			long v=e.get(1);
			e.remove(1);
			e.add(v);
		}else {
			long v=e.get(0);
			e.remove(0);
			e.add(v);
		}
		count++;
	}
}
while(q>0) {
	long v=s.nextLong();
	if(v<=count) {
		out.println(e1.get((int) (v-1)).a+" "+e1.get((int) (v-1)).b);
	}else {
		v-=count;
		long g=(long) (v%(n-1));
		if(g==0) {
			g=n-1;
		}
		out.println(e.get(0)+" "+e.get((int) g));
	}
	q--;
}
//ArrayList<Long> e=
		out.close();

	}
	static void dfs(int node) {
		vis[node]=1;
		for(int i=0;i<f[node].size();i++) {
			if(vis[f[node].get(i)]==0) {
				dfs(f[node].get(i));
			}
		}
	}
	static class pair{
		private long a;
		private long b;
		pair(long a,long b){
			this.a=a;
			this.b=b;
		}
	}
	public static long im(long a) {
		return binary_exponentiation_1(a,mod-2)%mod;
	}
	public static long binary_exponentiation_1(long a,long n) {
		long res=1;
		while(n>0) {
			if(n%2!=0) {
				res=((res)%(1000000007) * (a)%(1000000007))%(1000000007);
				n--;
			}else {
				a=((a)%(1000000007) *(a)%(1000000007))%(1000000007);
				n/=2;
			}
		}
		return (res)%(1000000007);
		
	}

	public static long[] fac=new long[100001];
	public static void find_factorial() {
		fac[0]=1;
		fac[1]=1;
		for(int i=2;i<=100000;i++) {
			fac[i]=(fac[i-1]*i)%(mod);
		}
	}
	static long c1=0;
	static long c2=0;
	static boolean bs1(long pos,long n,long x) {
		long l=0;
		long r=n;
		
		out:while(l<r) {
		long mid=(l+r)/2;
			if(mid<pos) {
//				if(l!=pos) {
//				c1++;}
				c1++;
				l=mid+1;
				
			}else {
				r=mid;
				c2++;
			}
		}
		if(c1<=x && c2<=(n-x)) {
//			return true;
//			System.out.println(c1+" "+c2);
		return true;
		}else {
			return false;
		}
		
	}
	static long v1=0;
	static long v2=0;
	static void bs2(long pos,long n,long x) {
		long left = 0, right = n;
		 
	    while(left < right) {
	        long middle = (left + right) / 2;
	        if (pos >= middle) {
	            if (pos != middle) v1++;
	            left = middle + 1;
	        }
	        else if (pos < middle){
	            v2++;
	            right = middle;
	        }
	    }
//		retur true
	    System.out.println(v1+" "+v2);
	}
	
	static long mod=1000000007;
	public static long GCD(long a,long b) {
		if(b==(long)0) {
			return a;
		}
		return GCD(b , a%b);
	}
	static long c=0;
	public static long res=1;
	public static void prime_factorization(long n) {
		for(long i=2;(long)(i*i)<=n;i++) {
			if(n%i==0) {
				long count=0;
				while(n%i==0) {
					count++;
					n/=i;
				}
//			prime_pairs.add(new pair(i,count));
//			res*=(count+1);
				c+=(count);
			}
		}
		if(n>1) {
//			prime_pairs.add(new pair(n,1));
//		res*=2;
			c+=(1);
		}
		
	}
	
		
	
}