#include <bits/stdc++.h>
using namespace std;
namespace IO {
const int sz = 1 << 15;
char inbuf[sz], outbuf[sz];
char *pinbuf = inbuf + sz;
char *poutbuf = outbuf;
inline char _getchar() {
  if (pinbuf == inbuf + sz) fread(inbuf, 1, sz, stdin), pinbuf = inbuf;
  return *(pinbuf++);
}
inline void _putchar(char x) {
  if (poutbuf == outbuf + sz) fwrite(outbuf, 1, sz, stdout), poutbuf = outbuf;
  *(poutbuf++) = x;
}
inline void flush() {
  if (poutbuf != outbuf)
    fwrite(outbuf, 1, poutbuf - outbuf, stdout), poutbuf = outbuf;
}
}  // namespace IO
inline int read() {
  int x = 0, p = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x * p;
}
const int Maxn = 100005;
int a[Maxn], n, q, Mx;
int ans1[Maxn], ans2[Maxn], ac;
int main() {
  deque<int> Q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    Mx = max(Mx, a[i]);
    Q.push_back(a[i]);
  }
  for (int i = 1;; i++) {
    int a = Q.front();
    Q.pop_front();
    int b = Q.front();
    Q.pop_front();
    if (a == Mx) {
      Q.push_front(b);
      Q.push_front(a);
      ac = i - 1;
      break;
    }
    ans1[i] = a;
    ans2[i] = b;
    Q.push_front(max(a, b));
    Q.push_back(min(a, b));
  }
  while (q--) {
    long long v;
    scanf("%I64d", &v);
    if (v <= ac) {
      printf("%d %d\n", ans1[v], ans2[v]);
    } else {
      v -= ac;
      v %= (n - 1);
      if (v == 0) v = n - 1;
      printf("%d %d\n", Mx, Q[v]);
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
namespace IO {
const int sz = 1 << 15;
char inbuf[sz], outbuf[sz];
char *pinbuf = inbuf + sz;
char *poutbuf = outbuf;
inline char _getchar() {
  if (pinbuf == inbuf + sz) fread(inbuf, 1, sz, stdin), pinbuf = inbuf;
  return *(pinbuf++);
}
inline void _putchar(char x) {
  if (poutbuf == outbuf + sz) fwrite(outbuf, 1, sz, stdout), poutbuf = outbuf;
  *(poutbuf++) = x;
}
inline void flush() {
  if (poutbuf != outbuf)
    fwrite(outbuf, 1, poutbuf - outbuf, stdout), poutbuf = outbuf;
}
}  // namespace IO
inline int read() {
  int x = 0, p = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x * p;
}
const int Maxn = 100005;
int a[Maxn], n, q, Mx;
int ans1[Maxn], ans2[Maxn], ac;
int main() {
  deque<int> Q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    Mx = max(Mx, a[i]);
    Q.push_back(a[i]);
  }
  for (int i = 1;; i++) {
    int a = Q.front();
    Q.pop_front();
    int b = Q.front();
    Q.pop_front();
    if (a == Mx) {
      Q.push_front(b);
      Q.push_front(a);
      ac = i - 1;
      break;
    }
    ans1[i] = a;
    ans2[i] = b;
    Q.push_front(max(a, b));
    Q.push_back(min(a, b));
  }
  while (q--) {
    long long v;
    scanf("%I64d", &v);
    if (v <= ac) {
      printf("%d %d\n", ans1[v], ans2[v]);
    } else {
      v -= ac;
      v %= (n - 1);
      if (v == 0) v = n - 1;
      printf("%d %d\n", Mx, Q[v]);
    }
  }
  return 0;
}

```