#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long n, t;
  cin >> n >> t;
  long long x;
  deque<long long> deq;
  long long m2 = -1;
  for (long i = 0; i < n; i++) {
    cin >> x;
    if (x > m2) m2 = x;
    deq.push_back(x);
  }
  if (t) {
    vector<long long> vec1;
    long long m = 0;
    for (long j = 0; j < t; j++) {
      cin >> x;
      if (x > m) m = x;
      vec1.push_back(x);
    }
    vector<pair<long, long> > vec2;
    long q = 0;
    int flag = 0;
    pair<long, long> y;
    long long temp;
    long z = 0;
    long i = 1;
    vector<pair<long, long> > vec3;
    vec3.push_back(make_pair(0, 0));
    for (; i <= m && deq[0] != m2; i++) {
      y.first = deq[0];
      y.second = deq[1];
      vec3.push_back(y);
      if (deq[0] > deq[1]) {
        temp = deq[1];
        deq.erase(deq.begin() + 1);
        deq.push_back(temp);
      } else {
        temp = deq[0];
        deq.pop_front();
        deq.push_back(temp);
      }
    }
    while (z < t) {
      if (vec1[z] < i) {
        vec2.push_back(vec3[vec1[z]]);
      } else {
        y.first = deq[0];
        y.second = deq[((vec1[z] - i) % (n - 1)) + 1];
        vec2.push_back(y);
      }
      z++;
    }
    for (long i = 0; i < t; i++) {
      cout << vec2[i].first << " " << vec2[i].second << endl;
    }
  }
}

### Prompt

Your challenge is to write 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long n, t;
  cin >> n >> t;
  long long x;
  deque<long long> deq;
  long long m2 = -1;
  for (long i = 0; i < n; i++) {
    cin >> x;
    if (x > m2) m2 = x;
    deq.push_back(x);
  }
  if (t) {
    vector<long long> vec1;
    long long m = 0;
    for (long j = 0; j < t; j++) {
      cin >> x;
      if (x > m) m = x;
      vec1.push_back(x);
    }
    vector<pair<long, long> > vec2;
    long q = 0;
    int flag = 0;
    pair<long, long> y;
    long long temp;
    long z = 0;
    long i = 1;
    vector<pair<long, long> > vec3;
    vec3.push_back(make_pair(0, 0));
    for (; i <= m && deq[0] != m2; i++) {
      y.first = deq[0];
      y.second = deq[1];
      vec3.push_back(y);
      if (deq[0] > deq[1]) {
        temp = deq[1];
        deq.erase(deq.begin() + 1);
        deq.push_back(temp);
      } else {
        temp = deq[0];
        deq.pop_front();
        deq.push_back(temp);
      }
    }
    while (z < t) {
      if (vec1[z] < i) {
        vec2.push_back(vec3[vec1[z]]);
      } else {
        y.first = deq[0];
        y.second = deq[((vec1[z] - i) % (n - 1)) + 1];
        vec2.push_back(y);
      }
      z++;
    }
    for (long i = 0; i < t; i++) {
      cout << vec2[i].first << " " << vec2[i].second << endl;
    }
  }
}

```