#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long int modInverse(long long int a) { return power(a, 1000000007 - 2); }
const int N = 500023;
bool vis[N];
vector<int> adj[N];
void solve() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  int pre_a[n + 1], pre_b[n + 1];
  int pos_max = -1;
  int mx = INT_MIN;
  for (long long int i = 0; i < n; i++) {
    if (a[i] > mx) {
      mx = a[i];
      pos_max = i;
    }
  }
  for (long long int i = 0; i < n; i++) {
    pre_a[i + 1] = a[0];
    pre_b[i + 1] = a[1];
    if (a[0] == mx) break;
    if (a[0] > a[1]) {
      a.push_back(a[1]);
      a.erase(a.begin() + 1);
    } else {
      a.push_back(a[0]);
      a.erase(a.begin());
    }
  }
  long long int p;
  for (long long int i = 0; i < q; i++) {
    cin >> p;
    if (p <= pos_max) {
      cout << pre_a[p] << " " << pre_b[p] << endl;
    } else {
      cout << mx << " " << a[(p - (pos_max + 1)) % (n - 1) + 1] << endl;
      ;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  int t = 0;
  while (t++ < T) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\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>
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long int modInverse(long long int a) { return power(a, 1000000007 - 2); }
const int N = 500023;
bool vis[N];
vector<int> adj[N];
void solve() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  int pre_a[n + 1], pre_b[n + 1];
  int pos_max = -1;
  int mx = INT_MIN;
  for (long long int i = 0; i < n; i++) {
    if (a[i] > mx) {
      mx = a[i];
      pos_max = i;
    }
  }
  for (long long int i = 0; i < n; i++) {
    pre_a[i + 1] = a[0];
    pre_b[i + 1] = a[1];
    if (a[0] == mx) break;
    if (a[0] > a[1]) {
      a.push_back(a[1]);
      a.erase(a.begin() + 1);
    } else {
      a.push_back(a[0]);
      a.erase(a.begin());
    }
  }
  long long int p;
  for (long long int i = 0; i < q; i++) {
    cin >> p;
    if (p <= pos_max) {
      cout << pre_a[p] << " " << pre_b[p] << endl;
    } else {
      cout << mx << " " << a[(p - (pos_max + 1)) % (n - 1) + 1] << endl;
      ;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  int t = 0;
  while (t++ < T) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

```