#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  while (T--) {
    long long int n, q;
    cin >> n >> q;
    std::vector<long long int> v;
    long long int arr[300010];
    std::vector<pair<long long int, long long int>> z;
    long long int maxa = -1;
    long long int ind;
    for (long long int i = 0; i < n; i = i + 1) {
      long long int a;
      cin >> a;
      arr[i] = a;
      v.push_back(a);
      if (maxa < a) {
        maxa = a;
        ind = i;
      }
    }
    std::vector<long long int> r;
    long long int start = 0, end = n - 1;
    while (arr[start] != maxa) {
      if (arr[start] < arr[start + 1]) {
        z.push_back(make_pair(arr[start], arr[start + 1]));
        end++;
        arr[end] = arr[start];
        start++;
      } else {
        z.push_back(make_pair(arr[start], arr[start + 1]));
        end++;
        arr[end] = arr[start + 1];
        arr[start + 1] = arr[start];
        start++;
      }
    }
    z.push_back(make_pair(arr[start], arr[start + 1]));
    start++;
    for (long long int i = start; i <= end; i++) {
      r.push_back(arr[i]);
    }
    while (q--) {
      long long int m;
      cin >> m;
      if (m <= ind + 1) {
        cout << z[m - 1].first << " " << z[m - 1].second << "\n";
      } else {
        long long int rem = m - ind;
        long long int tmp = r.size();
        long long int in = rem % tmp;
        if (in == 0) {
          in = tmp;
        }
        cout << maxa << " " << r[in - 1] << "\n";
      }
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  while (T--) {
    long long int n, q;
    cin >> n >> q;
    std::vector<long long int> v;
    long long int arr[300010];
    std::vector<pair<long long int, long long int>> z;
    long long int maxa = -1;
    long long int ind;
    for (long long int i = 0; i < n; i = i + 1) {
      long long int a;
      cin >> a;
      arr[i] = a;
      v.push_back(a);
      if (maxa < a) {
        maxa = a;
        ind = i;
      }
    }
    std::vector<long long int> r;
    long long int start = 0, end = n - 1;
    while (arr[start] != maxa) {
      if (arr[start] < arr[start + 1]) {
        z.push_back(make_pair(arr[start], arr[start + 1]));
        end++;
        arr[end] = arr[start];
        start++;
      } else {
        z.push_back(make_pair(arr[start], arr[start + 1]));
        end++;
        arr[end] = arr[start + 1];
        arr[start + 1] = arr[start];
        start++;
      }
    }
    z.push_back(make_pair(arr[start], arr[start + 1]));
    start++;
    for (long long int i = start; i <= end; i++) {
      r.push_back(arr[i]);
    }
    while (q--) {
      long long int m;
      cin >> m;
      if (m <= ind + 1) {
        cout << z[m - 1].first << " " << z[m - 1].second << "\n";
      } else {
        long long int rem = m - ind;
        long long int tmp = r.size();
        long long int in = rem % tmp;
        if (in == 0) {
          in = tmp;
        }
        cout << maxa << " " << r[in - 1] << "\n";
      }
    }
  }
  return 0;
}

```