#include <bits/stdc++.h>
using namespace std;
void canhazfast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T extgcd(T a, T b, T &x, T &y) {
  T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
  while (b) {
    T q = a / b;
    a %= b;
    swap(a, b);
    x0 -= q * x1;
    swap(x0, x1);
    y0 -= q * y1;
    swap(y0, y1);
  }
  x = x0;
  y = y0;
  return a;
}
int ctz(unsigned x) { return __builtin_ctz(x); }
int ctzll(unsigned long long x) { return __builtin_ctzll(x); }
int clz(unsigned x) { return __builtin_clz(x); }
int clzll(unsigned long long x) { return __builtin_clzll(x); }
int popcnt(unsigned x) { return __builtin_popcount(x); }
int popcntll(unsigned long long x) { return __builtin_popcountll(x); }
int bsr(unsigned x) { return 31 ^ clz(x); }
int bsrll(unsigned long long x) { return 63 ^ clzll(x); }
int a[2 * 100016];
pair<int, int> res[100016];
int main() {
  canhazfast();
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) {
    res[i] = {a[i], a[i + 1]};
    a[i + n] = min(a[i], a[i + 1]);
    a[i + 1] = max(a[i], a[i + 1]);
  }
  for (; q; --q) {
    long long m;
    cin >> m;
    --m;
    if (m < n)
      cout << res[m].first << ' ' << res[m].second << '\n';
    else
      cout << a[n] << ' ' << a[n + (m - 1) % (n - 1) + 1] << '\n';
  }
  return 0;
}

### Prompt

Generate 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;
void canhazfast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T extgcd(T a, T b, T &x, T &y) {
  T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
  while (b) {
    T q = a / b;
    a %= b;
    swap(a, b);
    x0 -= q * x1;
    swap(x0, x1);
    y0 -= q * y1;
    swap(y0, y1);
  }
  x = x0;
  y = y0;
  return a;
}
int ctz(unsigned x) { return __builtin_ctz(x); }
int ctzll(unsigned long long x) { return __builtin_ctzll(x); }
int clz(unsigned x) { return __builtin_clz(x); }
int clzll(unsigned long long x) { return __builtin_clzll(x); }
int popcnt(unsigned x) { return __builtin_popcount(x); }
int popcntll(unsigned long long x) { return __builtin_popcountll(x); }
int bsr(unsigned x) { return 31 ^ clz(x); }
int bsrll(unsigned long long x) { return 63 ^ clzll(x); }
int a[2 * 100016];
pair<int, int> res[100016];
int main() {
  canhazfast();
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) {
    res[i] = {a[i], a[i + 1]};
    a[i + n] = min(a[i], a[i + 1]);
    a[i + 1] = max(a[i], a[i + 1]);
  }
  for (; q; --q) {
    long long m;
    cin >> m;
    --m;
    if (m < n)
      cout << res[m].first << ' ' << res[m].second << '\n';
    else
      cout << a[n] << ' ' << a[n + (m - 1) % (n - 1) + 1] << '\n';
  }
  return 0;
}

```