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. 
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, q, i;
  cin >> n >> q;
  long long a[n + 1], b[q], a1[q], b1[q], mxa[n], mx, idx, aux[n];
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < q; i++) cin >> b[i];
  mx = a[0];
  mxa[0] = mx;
  idx = 0;
  for (i = 1; i < n; i++) {
    if (a[i] > mx) {
      mx = a[i];
      idx = i;
    }
    mxa[i] = mx;
  }
  long long g, h, itr = 0;
  g = a[0];
  h = a[1];
  for (i = idx + 1; i < n; i++) {
    aux[itr] = a[i];
    itr++;
  }
  for (i = 0; i < idx; i++) {
    h = a[i + 1];
    if (h > g) {
      aux[itr] = g;
      itr++;
      g = h;
      h = a[i + 2];
    } else {
      aux[itr] = h;
      itr++;
      h = a[i + 2];
    }
  }
  for (i = 0; i < q; i++) {
    if (b[i] > idx) {
      long long temp = b[i] - idx;
      a1[i] = mx;
      b1[i] = aux[(temp - 1) % (n - 1)];
    } else {
      a1[i] = mxa[b[i] - 1];
      b1[i] = a[b[i]];
    }
  }
  for (i = 0; i < q; i++) {
    cout << a1[i] << " " << b1[i] << endl;
  }
  return 0;
}
