#include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
using pa = pair<int, int>;
const double pi = acos(-1);
int size(int n) {
  int res = 1;
  while (res < n) res <<= 1;
  return res;
}
void fft(vector<complex<double>>& v) {
  int n = size(v.size());
  if (n > (int)v.size()) v.resize(n);
  for (int j = 1, i = 0; j < n - 1; j++) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(v[i], v[j]);
  }
  for (int m = 2; m <= n; m <<= 1) {
    double deg = (-1) * 2 * pi / m;
    complex<double> r(cos(deg), sin(deg));
    for (int i = 0; i < n; i += m) {
      complex<double> w(1, 0);
      for (int j = i, k = i + m / 2; k < i + m; j++, k++) {
        complex<double> t1 = v[j], t2 = w * v[k];
        v[j] = t1 + t2, v[k] = t1 - t2;
        w *= r;
      }
    }
  }
}
void ifft(vector<complex<double>>& v) {
  int n = size(v.size());
  if (n > (int)v.size()) v.resize(n);
  for (int j = 1, i = 0; j < n - 1; j++) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(v[i], v[j]);
  }
  for (int m = 2; m <= n; m <<= 1) {
    double deg = 2 * pi / m;
    complex<double> r(cos(deg), sin(deg));
    for (int i = 0; i < n; i += m) {
      complex<double> w(1, 0);
      for (int j = i, k = i + m / 2; k < i + m; j++, k++) {
        complex<double> t1 = v[j], t2 = w * v[k];
        v[j] = t1 + t2, v[k] = t1 - t2;
        w *= r;
      }
    }
  }
  for (int i = 0; i < n; ++i) v[i] *= 1.0 / n;
}
vector<int> convolution(const vector<int>& p, const vector<int>& q) {
  int n = size(p.size() + q.size());
  vector<complex<double>> P(n), Q(n);
  for (int i = 0; i < (int)p.size(); i++) P[i] = p[i];
  for (int i = 0; i < (int)q.size(); i++) Q[i] = q[i];
  fft(P);
  fft(Q);
  for (int i = 0; i < n; i++) P[i] /= Q[i];
  ifft(P);
  vector<int> res(n);
  for (int i = 0; i < n; i++) {
    res[i] = round(P[i].real());
  }
  return res;
}
int main() {
  int N, Q;
  cin >> N >> Q;
  deque<int> que;
  for (int i = 0; i < N; ++i) {
    int a;
    scanf("%d", &a);
    que.push_back(a);
  }
  vector<pair<int, int>> answers(3 * N);
  for (int i = 0; i < 3 * N; ++i) {
    int a = que.front();
    que.pop_front();
    int b = que.front();
    que.pop_front();
    answers[i] = make_pair(a, b);
    if (a > b) swap(a, b);
    que.push_front(b);
    que.push_back(a);
  }
  while (Q--) {
    long long int num;
    scanf("%lld", &num);
    num--;
    if (num >= 3 * N) {
      num %= (N - 1);
      num += 2 * (N - 1);
    }
    printf("%d %d\n", answers[num].first, answers[num].second);
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
#pragma warning(disable : 4996)
using pa = pair<int, int>;
const double pi = acos(-1);
int size(int n) {
  int res = 1;
  while (res < n) res <<= 1;
  return res;
}
void fft(vector<complex<double>>& v) {
  int n = size(v.size());
  if (n > (int)v.size()) v.resize(n);
  for (int j = 1, i = 0; j < n - 1; j++) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(v[i], v[j]);
  }
  for (int m = 2; m <= n; m <<= 1) {
    double deg = (-1) * 2 * pi / m;
    complex<double> r(cos(deg), sin(deg));
    for (int i = 0; i < n; i += m) {
      complex<double> w(1, 0);
      for (int j = i, k = i + m / 2; k < i + m; j++, k++) {
        complex<double> t1 = v[j], t2 = w * v[k];
        v[j] = t1 + t2, v[k] = t1 - t2;
        w *= r;
      }
    }
  }
}
void ifft(vector<complex<double>>& v) {
  int n = size(v.size());
  if (n > (int)v.size()) v.resize(n);
  for (int j = 1, i = 0; j < n - 1; j++) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(v[i], v[j]);
  }
  for (int m = 2; m <= n; m <<= 1) {
    double deg = 2 * pi / m;
    complex<double> r(cos(deg), sin(deg));
    for (int i = 0; i < n; i += m) {
      complex<double> w(1, 0);
      for (int j = i, k = i + m / 2; k < i + m; j++, k++) {
        complex<double> t1 = v[j], t2 = w * v[k];
        v[j] = t1 + t2, v[k] = t1 - t2;
        w *= r;
      }
    }
  }
  for (int i = 0; i < n; ++i) v[i] *= 1.0 / n;
}
vector<int> convolution(const vector<int>& p, const vector<int>& q) {
  int n = size(p.size() + q.size());
  vector<complex<double>> P(n), Q(n);
  for (int i = 0; i < (int)p.size(); i++) P[i] = p[i];
  for (int i = 0; i < (int)q.size(); i++) Q[i] = q[i];
  fft(P);
  fft(Q);
  for (int i = 0; i < n; i++) P[i] /= Q[i];
  ifft(P);
  vector<int> res(n);
  for (int i = 0; i < n; i++) {
    res[i] = round(P[i].real());
  }
  return res;
}
int main() {
  int N, Q;
  cin >> N >> Q;
  deque<int> que;
  for (int i = 0; i < N; ++i) {
    int a;
    scanf("%d", &a);
    que.push_back(a);
  }
  vector<pair<int, int>> answers(3 * N);
  for (int i = 0; i < 3 * N; ++i) {
    int a = que.front();
    que.pop_front();
    int b = que.front();
    que.pop_front();
    answers[i] = make_pair(a, b);
    if (a > b) swap(a, b);
    que.push_front(b);
    que.push_back(a);
  }
  while (Q--) {
    long long int num;
    scanf("%lld", &num);
    num--;
    if (num >= 3 * N) {
      num %= (N - 1);
      num += 2 * (N - 1);
    }
    printf("%d %d\n", answers[num].first, answers[num].second);
  }
  return 0;
}

```