#include <bits/stdc++.h>
template <typename T>
T Max(T x, T y) {
  return x > y ? x : y;
}
template <typename T>
T Min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T Abs(T x) {
  return x < 0 ? -x : x;
}
template <typename T>
T &read(T &r) {
  r = 0;
  bool w = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') w = ch == '-' ? 1 : 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    r = (r << 3) + (r << 1) + (ch ^ 48), ch = getchar();
  return r = w ? -r : r;
}
const int N = 1000010;
int n, a[N << 1], mx, p, b[N], ans1[N], ans2[N];
signed main() {
  read(n);
  int q;
  read(q);
  for (int i = 1; i <= n; ++i) b[i] = read(a[i]), mx = Max(mx, a[i]);
  p = 2;
  int pn = n;
  ans1[1] = a[1];
  ans2[1] = a[2];
  while (a[1] != mx) {
    if (a[1] > a[p])
      a[++pn] = a[p];
    else
      std::swap(a[1], a[p]), a[++pn] = a[p];
    ans1[p] = a[1];
    ans2[p] = a[p + 1];
    ++p;
  }
  --p;
  for (int i = 1; i <= q; ++i) {
    long long m;
    read(m);
    if (m < p)
      printf("%d %d\n", ans1[m], ans2[m]);
    else {
      printf("%d %d\n", mx,
             a[p +
               ((m - p + 1) % (n - 1) == 0 ? (n - 1) : (m - p + 1) % (n - 1))]);
    }
  }
  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>
template <typename T>
T Max(T x, T y) {
  return x > y ? x : y;
}
template <typename T>
T Min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T Abs(T x) {
  return x < 0 ? -x : x;
}
template <typename T>
T &read(T &r) {
  r = 0;
  bool w = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') w = ch == '-' ? 1 : 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    r = (r << 3) + (r << 1) + (ch ^ 48), ch = getchar();
  return r = w ? -r : r;
}
const int N = 1000010;
int n, a[N << 1], mx, p, b[N], ans1[N], ans2[N];
signed main() {
  read(n);
  int q;
  read(q);
  for (int i = 1; i <= n; ++i) b[i] = read(a[i]), mx = Max(mx, a[i]);
  p = 2;
  int pn = n;
  ans1[1] = a[1];
  ans2[1] = a[2];
  while (a[1] != mx) {
    if (a[1] > a[p])
      a[++pn] = a[p];
    else
      std::swap(a[1], a[p]), a[++pn] = a[p];
    ans1[p] = a[1];
    ans2[p] = a[p + 1];
    ++p;
  }
  --p;
  for (int i = 1; i <= q; ++i) {
    long long m;
    read(m);
    if (m < p)
      printf("%d %d\n", ans1[m], ans2[m]);
    else {
      printf("%d %d\n", mx,
             a[p +
               ((m - p + 1) % (n - 1) == 0 ? (n - 1) : (m - p + 1) % (n - 1))]);
    }
  }
  return 0;
}

```