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.lang.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class NitsLocal {
    static ArrayList<String> s1;
    static boolean[] prime;
    static int n = (int)1e7;
     static void sieve() {
Arrays.fill(prime	, true);
prime[0] = prime[1] = false;
for(int i = 2 ; i * i <= n ; ++i) {
if(prime[i]) {
for(int k = i * i; k<= n ; k+=i) {
prime[k] = false;
}
}
}

}
    public static void main(String[] args) {
        InputReader sc = new InputReader(System.in);
        
           prime  = new boolean[n + 1];
sieve();
                prime[1] = false;
               int x = sc.ni();
               int q = sc.ni();
               int []a = sc.nia(x);
               ArrayList<Integer> a1 = new ArrayList<>();
               int [][]res = new int[x+1][2];
               res[1][0] = a[0];
               res[1][1] = a[1];
               for(int i=1;i<x;i++)
               {
                   if(res[i][0] > res[i][1])
                   {
                       a1.add(res[i][1]);
                       res[i+1][0] = res[i][0];
                       res[i+1][1] = a[(i+1)%x];
                   }
                   else
                   {
                       a1.add(res[i][0]);
                       res[i+1][0] = a[i];
                       res[i+1][1] = a[(i+1)%x];
                   }
               }
               
               
                       int [][]res1 = new int[x-1][2];
                       res1[0][0] = res[x][0];
                       res1[0][1] = a1.get(0);
                       for(int i=1;i<x-1;i++)
                       {
                           res1[i][0] = res[x][0];
                           res1[i][1] = a1.get(i);
                       }
               
               while(q-- > 0)
               {
                   long r = sc.nl();
                   
                   if(r < (long)x)
                   {
                       
                   w.println(res[(int)r][0] + " " + res[(int)r][1]);
                   }
                   else
                   {
                         r -= (long)x;
                       r %= (long)(x-1);
                       
                     
                       w.println(res1[(int)r][0] + " " + res1[(int)r][1]);
                   }
               }
             
       w.close();
        
    }
    
   
  
   
    static String replace(String s,int a,int n)
    {
        char []c = s.toCharArray();
        for(int i=1;i<n;i+=2)
        {
            int num = (int) (c[i] - 48);
            num += a;
            num%=10;
            c[i] = (char) (num+48);
        }
        return new String(c);
    }
    static String move(String s,int h,int n)
    {
        h%=n;
        char []c = s.toCharArray();
        char []temp = new char[n];
        for(int i=0;i<n;i++)
        {
            temp[(i+h)%n] = c[i];
        }
        return new String(temp);
    }
    
     public static int ip(String s){
return Integer.parseInt(s);
}
     static class multipliers implements Comparator<Long>{
  
  public int compare(Long a,Long b) {
   if(a<b)
    return 1;
   else if(b<a)
    return -1;
   else
    return 0;
  }
 }
     static class multipliers1 implements Comparator<Student>{
  
  public int compare(Student a,Student b) {
   if(a.y<b.y)
    return 1;
   else if(b.y<a.y)
    return -1;
   else
   {
       if(a.id < b.id)
           return 1;
       else if(b.id<a.id)
           return -1;
       else
           return 0;
           
    //return 0;
   }
  }
 }
    // Java program to generate power set in 
// lexicographic order. 

    
    
    
      static class InputReader {
 
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
 
public InputReader(InputStream st) {
this.stream = st;
}
 
public int read() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
 
public int ni() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
 
public long nl() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
 
public int[] nia(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = ni();
}
return a;
}
 
public String rs() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
 
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
 
public boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
 
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
 
}





   
static PrintWriter w = new PrintWriter(System.out);
                  static class Student
    {
        int id;
        //int x;
       
        int y;
        //long z;
        
        Student(int id,int y)
        {
            this.id = id;
            //this.x = x;
            //this.s = s;
            this.y = y;
           // this.z = z;
            
            
            
        }
        
    }
}