#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e % 2)
    return b * power(b * b % m, (e - 1) / 2, m) % m;
  else
    return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e % 2)
    return b * power(b * b, (e - 1) / 2);
  else
    return power(b * b, e / 2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q, i;
  cin >> n >> q;
  long long m = 0;
  vector<long long> v(n);
  for (i = 0; i < n; i++) {
    cin >> v[i];
    m = max(m, v[i]);
  }
  for (i = 0; i < n; ++i) {
    if (v[i] == m) break;
  }
  vector<pair<long long, long long>> ans;
  long long moves = i;
  vector<long long> z;
  for (long long j = i; j < n; j++) z.push_back(v[j]);
  for (long long j = 0; j < i; j++) {
    ans.push_back({v[j], v[j + 1]});
    if (v[j] > v[j + 1]) {
      z.push_back(v[j + 1]);
      v[j + 1] = max(v[j], v[j + 1]);
    } else {
      z.push_back(v[j]);
      v[j + 1] = max(v[j], v[j + 1]);
    }
  }
  while (q--) {
    long long mi;
    cin >> mi;
    if (ans.size() >= mi)
      cout << ans[mi - 1].first << " " << ans[mi - 1].second << "\n";
    else {
      mi -= ans.size();
      long long temp = mi % (n - 1);
      if (temp == 0) temp = n - 1;
      cout << m << " " << z[temp] << "\n";
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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>
#pragma GCC optimize("Ofast")
using namespace std;
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e % 2)
    return b * power(b * b % m, (e - 1) / 2, m) % m;
  else
    return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e % 2)
    return b * power(b * b, (e - 1) / 2);
  else
    return power(b * b, e / 2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q, i;
  cin >> n >> q;
  long long m = 0;
  vector<long long> v(n);
  for (i = 0; i < n; i++) {
    cin >> v[i];
    m = max(m, v[i]);
  }
  for (i = 0; i < n; ++i) {
    if (v[i] == m) break;
  }
  vector<pair<long long, long long>> ans;
  long long moves = i;
  vector<long long> z;
  for (long long j = i; j < n; j++) z.push_back(v[j]);
  for (long long j = 0; j < i; j++) {
    ans.push_back({v[j], v[j + 1]});
    if (v[j] > v[j + 1]) {
      z.push_back(v[j + 1]);
      v[j + 1] = max(v[j], v[j + 1]);
    } else {
      z.push_back(v[j]);
      v[j + 1] = max(v[j], v[j + 1]);
    }
  }
  while (q--) {
    long long mi;
    cin >> mi;
    if (ans.size() >= mi)
      cout << ans[mi - 1].first << " " << ans[mi - 1].second << "\n";
    else {
      mi -= ans.size();
      long long temp = mi % (n - 1);
      if (temp == 0) temp = n - 1;
      cout << m << " " << z[temp] << "\n";
    }
  }
  return 0;
}

```