#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 1000;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
deque<long long> que;
long long a[maxn];
struct P {
  long long fi, se;
} ans1[maxn * 2], ans2[maxn * 2];
long long tot1 = 0, tot2 = 0;
int main(void) {
  cin.tie(0);
  std::ios::sync_with_stdio(false);
  long long n, q;
  cin >> n >> q;
  long long mx = -1;
  long long pos = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    que.push_back(a[i]);
    if (mx < a[i]) {
      mx = a[i];
      pos = i;
    }
  }
  for (long long i = 1; i <= 2 * n; i++) {
    long long one = que.front();
    que.pop_front();
    long long two = que.front();
    que.pop_front();
    ans1[++tot1].fi = one;
    ans1[tot1].se = two;
    if (one == mx) {
      ans2[++tot2].fi = one;
      ans2[tot2].se = two;
    }
    if (one > two) {
      que.push_front(one);
      que.push_back(two);
    } else {
      que.push_front(two);
      que.push_back(one);
    }
  }
  while (q--) {
    long long m;
    cin >> m;
    if (m <= pos) {
      cout << ans1[m].fi << " " << ans1[m].se << "\n";
    } else {
      long long cnt = m - pos;
      long long mod = cnt % (n - 1) + 1;
      cout << ans2[mod].fi << " " << ans2[mod].se << "\n";
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 maxn = 2e5 + 1000;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
deque<long long> que;
long long a[maxn];
struct P {
  long long fi, se;
} ans1[maxn * 2], ans2[maxn * 2];
long long tot1 = 0, tot2 = 0;
int main(void) {
  cin.tie(0);
  std::ios::sync_with_stdio(false);
  long long n, q;
  cin >> n >> q;
  long long mx = -1;
  long long pos = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    que.push_back(a[i]);
    if (mx < a[i]) {
      mx = a[i];
      pos = i;
    }
  }
  for (long long i = 1; i <= 2 * n; i++) {
    long long one = que.front();
    que.pop_front();
    long long two = que.front();
    que.pop_front();
    ans1[++tot1].fi = one;
    ans1[tot1].se = two;
    if (one == mx) {
      ans2[++tot2].fi = one;
      ans2[tot2].se = two;
    }
    if (one > two) {
      que.push_front(one);
      que.push_back(two);
    } else {
      que.push_front(two);
      que.push_back(one);
    }
  }
  while (q--) {
    long long m;
    cin >> m;
    if (m <= pos) {
      cout << ans1[m].fi << " " << ans1[m].se << "\n";
    } else {
      long long cnt = m - pos;
      long long mod = cnt % (n - 1) + 1;
      cout << ans2[mod].fi << " " << ans2[mod].se << "\n";
    }
  }
  return 0;
}

```