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.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;


public class q5 {
	
	public static void main(String[] args) throws IOException {
		
		Reader.init(System.in);
		PrintWriter out=new PrintWriter(System.out);
		int n=Reader.nextInt(),q=Reader.nextInt();
		LinkedList<Integer> ll=new LinkedList<Integer>();
		NoD[] query=new NoD[q];
		for(int i=0;i<n;i++) {
			ll.addLast(Reader.nextInt());
		}
		for(int i=0;i<q;i++) {
			query[i]=new NoD(Reader.nextLong(),i);
		}
		Arrays.sort(query,new Comparator<NoD>() {
			public int compare(NoD a, NoD b) {
				return Long.compare(a.a, b.a);
			}
		});
		int operations=0;
		int[] arr=new int[n];
		int[][] ans=new int[q][2];
		for(NoD a:query) {
			long v=a.a-operations;
			if(operations<n-1) {
				if(v==1) {
					int aa=ll.pollFirst();
					int b=ll.peekFirst();
					ll.addFirst(aa);
					ans[a.b][0]=aa;ans[a.b][1]=b;
				}
				else {
					while(v!=1 && operations<n-1) {
						operations++;
						v--;
						int aa=ll.pollFirst();
						int bb=ll.pollFirst();
						if(aa>bb) {
							ll.addFirst(aa);ll.addLast(bb);
						}
						else {
							ll.addLast(aa);ll.addFirst(bb);
						}
					}
					if(operations==n-1) {
						for(int i=0;i<n;i++) {
							arr[i]=ll.pollFirst();
						}
						int el=(int)(v%(n-1));
						if(el==0) el=n-1;
						ans[a.b][0]=arr[0];
						ans[a.b][1]=arr[el];
						
					}
					else {
						int aa=ll.pollFirst();
						int b=ll.peekFirst();
						ll.addFirst(aa);
						ans[a.b][0]=aa;ans[a.b][1]=b;
					}
				}
			}
			else {
				int el=(int)(v%(n-1));
				if(el==0) el=n-1;
				ans[a.b][0]=arr[0];
				ans[a.b][1]=arr[el];
			}
		}
		for(int i=0;i<q;i++) {
			out.println(ans[i][0]+" "+ans[i][1]);
		}
		
		out.flush();
		
	}
}
class NoD{
	int  b;
	long a;
	NoD(long aa,int bb){
		a=aa;b=bb;
	}
}



class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;
    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                     new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }
    /** get next word */
    static String nextLine() throws IOException{
    	return reader.readLine();
    }
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                   reader.readLine() );
        }
        return tokenizer.nextToken();
    }
    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }
    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}