#include <bits/stdc++.h>
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  int n, nquery;
  std::cin >> n >> nquery;
  std::vector<int> a(n);
  for (int& x : a) std::cin >> x;
  struct query {
    int qind;
    int64_t val;
    bool operator<(query q) const { return val > q.val; }
  };
  std::vector<query> qs(nquery);
  for (int qind = 0; qind < nquery; ++qind) {
    qs[qind].qind = qind;
    std::cin >> qs[qind].val;
    --qs[qind].val;
  }
  std::sort(begin(qs), end(qs));
  std::vector<std::pair<int, int>> out(nquery);
  int n_operations_done = 0;
  int const maxa = *std::max_element(begin(a), end(a));
  int start = 0;
  bool worked = false;
  auto const work = [&] {
    assert(not worked);
    assert(a[start] == maxa);
    for (auto& [qind, nop] : qs) {
      assert(nop >= n_operations_done);
      nop = (nop - n_operations_done) % (n - 1) + n_operations_done;
      assert(nop >= n_operations_done);
    }
    std::sort(begin(qs), end(qs));
    worked = true;
  };
  while (not qs.empty()) {
    auto const [qind, nop] = qs.back();
    qs.pop_back();
    while (n_operations_done < nop) {
      auto& x = a[start];
      auto& y = a[start + 1 == n ? 0 : start + 1];
      if (x > y) std::swap(x, y);
      start = start + 1 == n ? 0 : start + 1;
      ++n_operations_done;
      if (not worked and a[start] == maxa) {
        qs.push_back({qind, nop});
        work();
        goto cont_outer;
      }
    }
    assert(n_operations_done == nop);
    out[qind] = {a[start], a[start + 1 == n ? 0 : start + 1]};
  cont_outer:;
  }
  for (auto const [x, y] : out) std::cout << x << ' ' << y << '\n';
}

### Prompt

Please formulate 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>
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  int n, nquery;
  std::cin >> n >> nquery;
  std::vector<int> a(n);
  for (int& x : a) std::cin >> x;
  struct query {
    int qind;
    int64_t val;
    bool operator<(query q) const { return val > q.val; }
  };
  std::vector<query> qs(nquery);
  for (int qind = 0; qind < nquery; ++qind) {
    qs[qind].qind = qind;
    std::cin >> qs[qind].val;
    --qs[qind].val;
  }
  std::sort(begin(qs), end(qs));
  std::vector<std::pair<int, int>> out(nquery);
  int n_operations_done = 0;
  int const maxa = *std::max_element(begin(a), end(a));
  int start = 0;
  bool worked = false;
  auto const work = [&] {
    assert(not worked);
    assert(a[start] == maxa);
    for (auto& [qind, nop] : qs) {
      assert(nop >= n_operations_done);
      nop = (nop - n_operations_done) % (n - 1) + n_operations_done;
      assert(nop >= n_operations_done);
    }
    std::sort(begin(qs), end(qs));
    worked = true;
  };
  while (not qs.empty()) {
    auto const [qind, nop] = qs.back();
    qs.pop_back();
    while (n_operations_done < nop) {
      auto& x = a[start];
      auto& y = a[start + 1 == n ? 0 : start + 1];
      if (x > y) std::swap(x, y);
      start = start + 1 == n ? 0 : start + 1;
      ++n_operations_done;
      if (not worked and a[start] == maxa) {
        qs.push_back({qind, nop});
        work();
        goto cont_outer;
      }
    }
    assert(n_operations_done == nop);
    out[qind] = {a[start], a[start + 1 == n ? 0 : start + 1]};
  cont_outer:;
  }
  for (auto const [x, y] : out) std::cout << x << ' ' << y << '\n';
}

```