#include <bits/stdc++.h>
using namespace std;
const int mx1 = 15;
const int mx2 = 505;
const int mx3 = 1005;
const int mx4 = 10005;
const int mx5 = 100005;
const int mx6 = 1000005;
int _toggle(int N, int pos) { return N = N ^ (1 << pos); }
int _set(int N, int pos) { return N = N | (1 << pos); }
int _reset(int N, int pos) { return N = N & ~(1 << pos); }
bool _check(int N, int pos) { return (bool)(N & (1 << pos)); }
bool _upper(char a) { return a >= 'A' && a <= 'Z'; }
bool _lower(char a) { return a >= 'a' && a <= 'z'; }
bool _digit(char a) { return a >= '0' && a <= '9'; }
int dx[] = {1, -1, 0, 0, -1, -1, 1, 1};
int dy[] = {0, 0, 1, -1, -1, 1, -1, 1};
int n, q, a, mx, ans[mx5][2];
deque<int> Q;
void solve() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    mx = max(a, mx);
    Q.push_back(a);
  }
  if (q == 0) {
    cout << '\n';
    return;
  }
  vector<int> stt;
  int pre, f = 0;
  for (int i = 1; i <= n; i++) {
    int A = Q.front();
    Q.pop_front();
    int B = Q.front();
    Q.pop_front();
    if (A == mx && !f) {
      f = 1;
      pre = i - 1;
      stt.push_back(B);
      while (!Q.empty()) {
        stt.push_back(Q.front());
        Q.pop_front();
      }
      break;
    }
    if (A > B) {
      Q.push_front(A);
      Q.push_back(B);
    } else {
      Q.push_front(B);
      Q.push_back(A);
    }
    ans[i][0] = A, ans[i][1] = B;
  }
  while (q--) {
    long long x;
    cin >> x;
    if (x <= pre) {
      cout << ans[x][0] << ' ' << ans[x][1] << '\n';
    } else {
      x -= pre;
      x %= (n - 1);
      if (!x) {
        x = n - 1;
      }
      x--;
      cout << mx << ' ' << stt[x] << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  while (tc--) {
    solve();
  }
}

### Prompt

Your challenge is to write a cpp solution to 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>
using namespace std;
const int mx1 = 15;
const int mx2 = 505;
const int mx3 = 1005;
const int mx4 = 10005;
const int mx5 = 100005;
const int mx6 = 1000005;
int _toggle(int N, int pos) { return N = N ^ (1 << pos); }
int _set(int N, int pos) { return N = N | (1 << pos); }
int _reset(int N, int pos) { return N = N & ~(1 << pos); }
bool _check(int N, int pos) { return (bool)(N & (1 << pos)); }
bool _upper(char a) { return a >= 'A' && a <= 'Z'; }
bool _lower(char a) { return a >= 'a' && a <= 'z'; }
bool _digit(char a) { return a >= '0' && a <= '9'; }
int dx[] = {1, -1, 0, 0, -1, -1, 1, 1};
int dy[] = {0, 0, 1, -1, -1, 1, -1, 1};
int n, q, a, mx, ans[mx5][2];
deque<int> Q;
void solve() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    mx = max(a, mx);
    Q.push_back(a);
  }
  if (q == 0) {
    cout << '\n';
    return;
  }
  vector<int> stt;
  int pre, f = 0;
  for (int i = 1; i <= n; i++) {
    int A = Q.front();
    Q.pop_front();
    int B = Q.front();
    Q.pop_front();
    if (A == mx && !f) {
      f = 1;
      pre = i - 1;
      stt.push_back(B);
      while (!Q.empty()) {
        stt.push_back(Q.front());
        Q.pop_front();
      }
      break;
    }
    if (A > B) {
      Q.push_front(A);
      Q.push_back(B);
    } else {
      Q.push_front(B);
      Q.push_back(A);
    }
    ans[i][0] = A, ans[i][1] = B;
  }
  while (q--) {
    long long x;
    cin >> x;
    if (x <= pre) {
      cout << ans[x][0] << ' ' << ans[x][1] << '\n';
    } else {
      x -= pre;
      x %= (n - 1);
      if (!x) {
        x = n - 1;
      }
      x--;
      cout << mx << ' ' << stt[x] << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  while (tc--) {
    solve();
  }
}

```