#include <bits/stdc++.h>
using namespace std;
long long a[205000] = {0}, b[205000] = {0};
long long zuida[205000] = {0};
long long ci, maxx = -12345;
int ask, n;
int h, t;
int main() {
  int bj;
  cin >> n >> ask;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    b[i] = a[i];
    if (a[i] > maxx) {
      maxx = a[i];
      bj = i;
    }
  }
  zuida[1] = a[1];
  for (int i = 2; i <= n; i++) {
    if (a[i] > zuida[i - 1]) {
      zuida[i] = a[i];
    } else
      zuida[i] = zuida[i - 1];
  }
  h = 1, t = n + 1;
  long long pd;
  pd = bj - 2 + 1;
  for (int i = 1; i <= pd; i++) {
    if (b[h] > b[h + 1]) {
      b[t] = b[h + 1];
      b[h + 1] = b[h];
      h++;
      t++;
    } else {
      b[t] = b[h];
      h++;
      t++;
    }
  }
  if (ask == 0) {
    return 0;
  }
  for (int i = 1; i <= ask; i++) {
    scanf("%lld", &ci);
    if (ci == 0) {
      cout << 0 << " " << 0 << endl;
      continue;
    }
    if (ci <= pd) {
      printf("%lld %lld\n", zuida[ci], a[ci + 1]);
    } else {
      long long yu = (ci - pd) % (n - 1);
      if (yu == 0)
        printf("%lld %lld\n", maxx, b[bj + n - 1]);
      else
        printf("%lld %lld\n", maxx, b[bj + yu]);
    }
  }
  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>
using namespace std;
long long a[205000] = {0}, b[205000] = {0};
long long zuida[205000] = {0};
long long ci, maxx = -12345;
int ask, n;
int h, t;
int main() {
  int bj;
  cin >> n >> ask;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    b[i] = a[i];
    if (a[i] > maxx) {
      maxx = a[i];
      bj = i;
    }
  }
  zuida[1] = a[1];
  for (int i = 2; i <= n; i++) {
    if (a[i] > zuida[i - 1]) {
      zuida[i] = a[i];
    } else
      zuida[i] = zuida[i - 1];
  }
  h = 1, t = n + 1;
  long long pd;
  pd = bj - 2 + 1;
  for (int i = 1; i <= pd; i++) {
    if (b[h] > b[h + 1]) {
      b[t] = b[h + 1];
      b[h + 1] = b[h];
      h++;
      t++;
    } else {
      b[t] = b[h];
      h++;
      t++;
    }
  }
  if (ask == 0) {
    return 0;
  }
  for (int i = 1; i <= ask; i++) {
    scanf("%lld", &ci);
    if (ci == 0) {
      cout << 0 << " " << 0 << endl;
      continue;
    }
    if (ci <= pd) {
      printf("%lld %lld\n", zuida[ci], a[ci + 1]);
    } else {
      long long yu = (ci - pd) % (n - 1);
      if (yu == 0)
        printf("%lld %lld\n", maxx, b[bj + n - 1]);
      else
        printf("%lld %lld\n", maxx, b[bj + yu]);
    }
  }
  return 0;
}

```