#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
template <class C>
void mini(C &a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C &a4, C b4) {
  a4 = max(a4, b4);
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
long long n, q;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(11);
  if (0) cout << fixed << setprecision(6);
  cin >> n >> q;
  vector<pair<long long, long long> > res;
  deque<long long> x;
  long long ma = -1;
  for (long long i = (0); i <= ((long long)(n)-1); i++) {
    long long pom;
    cin >> pom;
    maxi(ma, pom);
    x.push_back(pom);
  }
  while (x[0] != ma) {
    res.push_back({x[0], x[1]});
    long long pom = max(x[0], x[1]);
    long long pom2 = min(x[0], x[1]);
    x.pop_front();
    x.pop_front();
    x.push_front(pom);
    x.push_back(pom2);
  }
  for (long long i = (0); i <= ((long long)(q)-1); i++) {
    long long pom;
    cin >> pom;
    pom--;
    if (pom < ((long long)(res).size())) {
      cout << res[pom].first << " " << res[pom].second << "\n";
    } else {
      pom -= ((long long)(res).size());
      pom %= ((long long)(x).size()) - 1;
      cout << x[0] << " " << x[pom + 1] << "\n";
    }
  }
}

### 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>
#pragma GCC optimize("O3")
using namespace std;
template <class C>
void mini(C &a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C &a4, C b4) {
  a4 = max(a4, b4);
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
long long n, q;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(11);
  if (0) cout << fixed << setprecision(6);
  cin >> n >> q;
  vector<pair<long long, long long> > res;
  deque<long long> x;
  long long ma = -1;
  for (long long i = (0); i <= ((long long)(n)-1); i++) {
    long long pom;
    cin >> pom;
    maxi(ma, pom);
    x.push_back(pom);
  }
  while (x[0] != ma) {
    res.push_back({x[0], x[1]});
    long long pom = max(x[0], x[1]);
    long long pom2 = min(x[0], x[1]);
    x.pop_front();
    x.pop_front();
    x.push_front(pom);
    x.push_back(pom2);
  }
  for (long long i = (0); i <= ((long long)(q)-1); i++) {
    long long pom;
    cin >> pom;
    pom--;
    if (pom < ((long long)(res).size())) {
      cout << res[pom].first << " " << res[pom].second << "\n";
    } else {
      pom -= ((long long)(res).size());
      pom %= ((long long)(x).size()) - 1;
      cout << x[0] << " " << x[pom + 1] << "\n";
    }
  }
}

```