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. 
// Working program using Reader Class 
import java.io.*; 
import java.io.FileInputStream; 
import java.io.IOException; 
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*; 

public class Main 
{ 
	static class Reader 
	{ 
		final private int BUFFER_SIZE = 1 << 16; 
		private DataInputStream din; 
		private byte[] buffer; 
		private int bufferPointer, bytesRead; 

		public Reader() 
		{ 
			din = new DataInputStream(System.in); 
			buffer = new byte[BUFFER_SIZE]; 
			bufferPointer = bytesRead = 0; 
		} 

		public Reader(String file_name) throws IOException 
		{ 
			din = new DataInputStream(new FileInputStream(file_name)); 
			buffer = new byte[BUFFER_SIZE]; 
			bufferPointer = bytesRead = 0; 
		} 

		public String readLine() throws IOException 
		{ 
			byte[] buf = new byte[64]; // line length 
			int cnt = 0, c; 
			while ((c = read()) != -1) 
			{ 
				if (c == '\n') 
					break; 
				buf[cnt++] = (byte) c; 
			} 
			return new String(buf, 0, cnt); 
		} 

		public int nextInt() throws IOException 
		{ 
			int ret = 0; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
			do
			{ 
				ret = ret * 10 + c - '0'; 
			} while ((c = read()) >= '0' && c <= '9'); 

			if (neg) 
				return -ret; 
			return ret; 
		} 

		public long nextLong() throws IOException 
		{ 
			long ret = 0; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
			do { 
				ret = ret * 10 + c - '0'; 
			} 
			while ((c = read()) >= '0' && c <= '9'); 
			if (neg) 
				return -ret; 
			return ret; 
		} 

		public double nextDouble() throws IOException 
		{ 
			double ret = 0, div = 1; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 

			do { 
				ret = ret * 10 + c - '0'; 
			} 
			while ((c = read()) >= '0' && c <= '9'); 

			if (c == '.') 
			{ 
				while ((c = read()) >= '0' && c <= '9') 
				{ 
					ret += (c - '0') / (div *= 10); 
				} 
			} 

			if (neg) 
				return -ret; 
			return ret; 
		} 

		private void fillBuffer() throws IOException 
		{ 
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
			if (bytesRead == -1) 
				buffer[0] = -1; 
		} 

		private byte read() throws IOException 
		{ 
			if (bufferPointer == bytesRead) 
				fillBuffer(); 
			return buffer[bufferPointer++]; 
		} 

		public void close() throws IOException 
		{ 
			if (din == null) 
				return; 
			din.close(); 
		} 
	} 
	 
	public static int gcd(int a, int b) 
    { 
        // Everything divides 0  
        if (a == 0 || b == 0) 
           return 0; 
       
        // base case 
        if (a == b) 
            return a; 
       
        // a is greater 
        if (a > b) 
            return gcd(a-b, b); 
        return gcd(a, b-a); 
    }
	
	public static class Node{
		int val1,val2;
		Node(int v,int i){
			val1=v;
			val2=i;
		}
	}
	
	public static void main(String[] args) throws IOException 
	{ 
		//Reader scan=new Reader(); 
		Reader scan=new Reader();
		PrintWriter out=new PrintWriter(System.out);
		int n=scan.nextInt();
		int q=scan.nextInt();
		ArrayDeque<Integer> adq=new ArrayDeque<Integer>();
		//int maxi=-1;
		int max=Integer.MIN_VALUE;
		for(int i=0;i<n;i++) {
			int x=scan.nextInt();
			max=Math.max(max, x);
			adq.addLast(x);
		}
		ArrayList<Node> al=new ArrayList<Node>();
		while(adq.getFirst()!=max) {
			int x=adq.pollFirst();
			int y=adq.pollFirst();
			al.add(new Node(x,y));
			if(x>y) {
				adq.addFirst(x);
				adq.addLast(y);
			}
			else {
				adq.addFirst(y);
				adq.addLast(x);
			}
		}
		adq.pollFirst();
		int[] ans=new int[n-1];
		int i=0;
		while(!adq.isEmpty()) {
			ans[i]=adq.pollFirst();
			i++;
		}
		while(q-->0) {
			long m=scan.nextLong();
			if(al.size()>=m) {
				out.print(al.get((int)(m-1)).val1+" "+al.get((int)(m-1)).val2);
				out.println();
			}
			else {
				int z=(int)((m-(long)al.size())%(long)(n-1));
				if(z==0) {
					z=n-1;
				}
				out.print(max+" "+ans[z-1]);
				out.print("\n");
			}
		}
		out.close();
	} 
}