#include <bits/stdc++.h>
using namespace std;
inline pair<vector<int>, vector<pair<int, int>>> get_new_arr(vector<int> &arr,
                                                             int pos) {
  queue<int> fifo;
  vector<pair<int, int>> ans;
  if (pos == 0) {
    vector<int> new_arr(arr.size() - 1);
    for (int i = 1; i < arr.size(); i++) {
      new_arr[i - 1] = arr[i];
    }
    return {new_arr, ans};
  }
  for (int i = pos + 1; i < arr.size(); i++) {
    fifo.push(arr[i]);
  }
  int big = arr[0];
  for (int i = 1; i < pos; i++) {
    if (arr[i] <= big) {
      fifo.push(arr[i]);
      ans.push_back({big, arr[i]});
    } else {
      fifo.push(big);
      ans.push_back({big, arr[i]});
      big = arr[i];
    }
  }
  if (pos > 0) {
    ans.push_back({big, arr[pos]});
    fifo.push(big);
  }
  vector<int> new_arr;
  while (!fifo.empty()) {
    new_arr.push_back(fifo.front());
    fifo.pop();
  }
  pair<vector<int>, vector<pair<int, int>>> result = {new_arr, ans};
  return result;
}
int main() {
  long long n, q;
  scanf("%lld %lld", &n, &q);
  vector<int> a(n);
  int big = -1, pos, aux;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (big < a[i]) {
      big = a[i];
      pos = i;
    }
  }
  pair<vector<int>, vector<pair<int, int>>> result = get_new_arr(a, pos);
  vector<int> arr = result.first;
  vector<pair<int, int>> pre_ans = result.second;
  long long query;
  pair<int, int> p;
  while (q--) {
    scanf("%lld", &query);
    if (query <= pos) {
      p = pre_ans[query - 1];
    } else {
      long long x = query;
      query = (query - pos - 1) % (n - 1);
      if (arr[query] == 0) {
        printf("Aqui %lld %lld\n", x, query);
      }
      p = {a[pos], arr[query]};
    }
    printf("%d %d\n", p.first, p.second);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline pair<vector<int>, vector<pair<int, int>>> get_new_arr(vector<int> &arr,
                                                             int pos) {
  queue<int> fifo;
  vector<pair<int, int>> ans;
  if (pos == 0) {
    vector<int> new_arr(arr.size() - 1);
    for (int i = 1; i < arr.size(); i++) {
      new_arr[i - 1] = arr[i];
    }
    return {new_arr, ans};
  }
  for (int i = pos + 1; i < arr.size(); i++) {
    fifo.push(arr[i]);
  }
  int big = arr[0];
  for (int i = 1; i < pos; i++) {
    if (arr[i] <= big) {
      fifo.push(arr[i]);
      ans.push_back({big, arr[i]});
    } else {
      fifo.push(big);
      ans.push_back({big, arr[i]});
      big = arr[i];
    }
  }
  if (pos > 0) {
    ans.push_back({big, arr[pos]});
    fifo.push(big);
  }
  vector<int> new_arr;
  while (!fifo.empty()) {
    new_arr.push_back(fifo.front());
    fifo.pop();
  }
  pair<vector<int>, vector<pair<int, int>>> result = {new_arr, ans};
  return result;
}
int main() {
  long long n, q;
  scanf("%lld %lld", &n, &q);
  vector<int> a(n);
  int big = -1, pos, aux;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (big < a[i]) {
      big = a[i];
      pos = i;
    }
  }
  pair<vector<int>, vector<pair<int, int>>> result = get_new_arr(a, pos);
  vector<int> arr = result.first;
  vector<pair<int, int>> pre_ans = result.second;
  long long query;
  pair<int, int> p;
  while (q--) {
    scanf("%lld", &query);
    if (query <= pos) {
      p = pre_ans[query - 1];
    } else {
      long long x = query;
      query = (query - pos - 1) % (n - 1);
      if (arr[query] == 0) {
        printf("Aqui %lld %lld\n", x, query);
      }
      p = {a[pos], arr[query]};
    }
    printf("%d %d\n", p.first, p.second);
  }
  return 0;
}

```