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>
const long long int INF = (long long int)1e9;
const double EPS = 0.000000000001;
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  long long int n, q;
  cin >> n >> q;
  deque<long long int> deq;
  long long int max_n = -1;
  for (long long int i = 0; i < n; ++i) {
    long long int temp;
    cin >> temp;
    max_n = max(max_n, temp);
    deq.push_back(temp);
  }
  vector<pair<long long int, long long int> > operations_prev;
  while (deq.front() != max_n) {
    long long int a, b;
    a = deq.front();
    deq.pop_front();
    b = deq.front();
    deq.pop_front();
    operations_prev.push_back({a, b});
    if (a > b) {
      deq.push_front(a);
      deq.push_back(b);
    } else {
      deq.push_front(b);
      deq.push_back(a);
    }
  }
  vector<long long int> arr(n - 1);
  deq.pop_front();
  for (long long int i = 0; i < n - 1; ++i) {
    arr[i] = deq.front();
    deq.pop_front();
  }
  for (long long int i = 0; i < q; ++i) {
    long long int temp;
    cin >> temp;
    if (temp <= operations_prev.size()) {
      cout << operations_prev[temp - 1].first << ' '
           << operations_prev[temp - 1].second << '\n';
    } else {
      temp -= operations_prev.size() + 1;
      temp = (temp % (n - 1));
      cout << max_n << ' ' << arr[temp] << '\n';
    }
  }
}
