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. 
#    !/usr/bin/env python3
#    encoding: UTF-8
#    Modified: <22/Jun/2019 12:28:28 AM>


#    ✪ H4WK3yE乡
#    Mohd. Farhan Tahir
#    Indian Institute Of Information Technology (IIIT), Gwalior


import sys
import os
from io import IOBase, BytesIO


def main():
    from collections import deque
    n, tc = get_ints()
    arr = get_array()
    mx = max(arr)
    q = deque()
    for i in range(n):
        q.append(arr[i])
    query = 1
    dp = [0] * 10**6
    while q[0] != mx:
        f = q.popleft()
        s = q.popleft()
        if (f > s):
            q.appendleft(f)
            q.append(s)
        else:
            q.appendleft(s)
            q.append(f)
        dp[query] = (f, s)
        query += 1

    curr = 0
    c2 = query
    while curr != n - 1:
        #print(q, c2)
        f = q.popleft()
        s = q.popleft()
        q.appendleft(f)
        q.append(s)
        dp[c2] = (f, s)
        curr += 1
        c2 += 1
    curr = 0
    cycle = c2 - query
    while curr != n - 1:
        #print(q, c2)
        f = q.popleft()
        s = q.popleft()
        q.appendleft(f)
        q.append(s)
        dp[c2] = (f, s)
        curr += 1
        c2 += 1
    # print(dp[:10])
    #print(cycle, query, c2)
    # print(dp[:10])
    # print(c2)
    for _ in range(tc):
        j = int(input())
        if j < c2:
            print(*dp[j])
        else:
            j %= cycle
            j += cycle
            #print(j, "j")
            print(*dp[j])

    # print(dp)


def get_array(): return list(map(int, sys.stdin.readline().split()))


def get_ints(): return map(int, sys.stdin.readline().split())


def input(): return sys.stdin.readline().strip()


if __name__ == "__main__":
    main()
