#include <bits/stdc++.h>
using namespace std;
const int QQ = 3e5 + 10;
const int N = 1e5;
const int module = 1e9 + 7;
long long rand(long long l, long long r) {
  return 1LL * (1LL * rand() * RAND_MAX + rand()) % (r - l + 1) + l;
}
void doc() {}
void xuli(int TestCase) {
  cout << "Case #" << TestCase << '\n';
  double Pi = atan(1) * 4;
  double Res = 0;
  double R, Start, End;
  cin >> R >> Start >> End;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    double l, r, H;
    cin >> l >> r >> H;
    if (H >= R) {
      Res += (r - Start - R);
      Res += 2.0 * (H - R);
      Res += R * Pi;
      Start = 1.0 * r + R;
    } else {
      double Phi = Pi - 2.0 * asin(double(R - H) / R);
      double dist = sqrt(2 * R * H - H * H);
      Res += (r - Start - dist);
      Res += R * Phi;
      Start = r + dist;
    }
  }
  Res += End - Start;
  cout << fixed << setprecision(15) << Res << '\n';
}
void solve() {
  int n, Q;
  cin >> n >> Q;
  int mx = 0;
  int a[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    mx = max(mx, a[i]);
  }
  vector<int> Pos;
  int id = 0;
  pair<int, int> Res[n + 1];
  int L = 1;
  int R = 2;
  int KT[n + 1];
  memset(KT, 0, sizeof(KT));
  while (a[L] != mx) {
    id++;
    Res[id] = pair<int, int>(a[L], a[R]);
    if (max(a[L], a[R]) == a[L]) {
      R++;
      Pos.push_back(R - 1);
      KT[R - 1] = 1;
    } else {
      Pos.push_back(L);
      KT[L] = 1;
      L = R;
      R++;
    }
  }
  int m = 0;
  int b[n + 1];
  for (int i = 1; i <= n; i++) {
    if (KT[i] == 0) {
      b[++m] = a[i];
    }
  }
  for (int x : Pos) {
    b[++m] = a[x];
  }
  m = n - 1;
  while (Q--) {
    long long VT;
    cin >> VT;
    if (VT <= id) {
      cout << Res[VT].first << " " << Res[VT].second << '\n';
      continue;
    }
    VT -= (id + 1);
    VT %= (n - 1);
    cout << b[1] << " " << b[2 + VT] << '\n';
  }
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  doc();
  solve();
  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;
const int QQ = 3e5 + 10;
const int N = 1e5;
const int module = 1e9 + 7;
long long rand(long long l, long long r) {
  return 1LL * (1LL * rand() * RAND_MAX + rand()) % (r - l + 1) + l;
}
void doc() {}
void xuli(int TestCase) {
  cout << "Case #" << TestCase << '\n';
  double Pi = atan(1) * 4;
  double Res = 0;
  double R, Start, End;
  cin >> R >> Start >> End;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    double l, r, H;
    cin >> l >> r >> H;
    if (H >= R) {
      Res += (r - Start - R);
      Res += 2.0 * (H - R);
      Res += R * Pi;
      Start = 1.0 * r + R;
    } else {
      double Phi = Pi - 2.0 * asin(double(R - H) / R);
      double dist = sqrt(2 * R * H - H * H);
      Res += (r - Start - dist);
      Res += R * Phi;
      Start = r + dist;
    }
  }
  Res += End - Start;
  cout << fixed << setprecision(15) << Res << '\n';
}
void solve() {
  int n, Q;
  cin >> n >> Q;
  int mx = 0;
  int a[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    mx = max(mx, a[i]);
  }
  vector<int> Pos;
  int id = 0;
  pair<int, int> Res[n + 1];
  int L = 1;
  int R = 2;
  int KT[n + 1];
  memset(KT, 0, sizeof(KT));
  while (a[L] != mx) {
    id++;
    Res[id] = pair<int, int>(a[L], a[R]);
    if (max(a[L], a[R]) == a[L]) {
      R++;
      Pos.push_back(R - 1);
      KT[R - 1] = 1;
    } else {
      Pos.push_back(L);
      KT[L] = 1;
      L = R;
      R++;
    }
  }
  int m = 0;
  int b[n + 1];
  for (int i = 1; i <= n; i++) {
    if (KT[i] == 0) {
      b[++m] = a[i];
    }
  }
  for (int x : Pos) {
    b[++m] = a[x];
  }
  m = n - 1;
  while (Q--) {
    long long VT;
    cin >> VT;
    if (VT <= id) {
      cout << Res[VT].first << " " << Res[VT].second << '\n';
      continue;
    }
    VT -= (id + 1);
    VT %= (n - 1);
    cout << b[1] << " " << b[2 + VT] << '\n';
  }
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  doc();
  solve();
  return 0;
}

```