#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long maxn = 500005;
const long long mm = 1e+18;
long long power(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long ncr(long long n, long long k) {
  long long C[k + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, k); j > 0; j--) C[j] = (C[j] + C[j - 1]) % mod;
  }
  return C[k];
}
void solve() {
  int n, q1;
  cin >> n >> q1;
  int arr[n];
  int mx = -1;
  deque<int> q;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    q.push_back(arr[i]);
    if (arr[i] > mx) {
      mx = arr[i];
    }
  }
  vector<pair<int, int>> v;
  v.push_back({0, 0});
  long long tot = 0;
  while (q.front() != mx) {
    int x1 = q.front();
    q.pop_front();
    int x2 = q.front();
    q.pop_front();
    if (x1 > x2) {
      q.push_front(x1);
      q.push_back(x2);
    } else {
      q.push_front(x2);
      q.push_back(x1);
    }
    tot++;
    v.push_back({x1, x2});
  }
  vector<int> vv;
  q.pop_front();
  while (q.size()) {
    vv.push_back(q.front());
    q.pop_front();
  }
  while (q1--) {
    long long x;
    cin >> x;
    if (x <= tot) {
      cout << v[x].first << " " << v[x].second << "\n";
    } else {
      cout << mx << " ";
      long long xx = (x - tot - 1) % (n - 1);
      cout << vv[xx] << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 long long mod = 1000000007;
const long long maxn = 500005;
const long long mm = 1e+18;
long long power(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long ncr(long long n, long long k) {
  long long C[k + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, k); j > 0; j--) C[j] = (C[j] + C[j - 1]) % mod;
  }
  return C[k];
}
void solve() {
  int n, q1;
  cin >> n >> q1;
  int arr[n];
  int mx = -1;
  deque<int> q;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    q.push_back(arr[i]);
    if (arr[i] > mx) {
      mx = arr[i];
    }
  }
  vector<pair<int, int>> v;
  v.push_back({0, 0});
  long long tot = 0;
  while (q.front() != mx) {
    int x1 = q.front();
    q.pop_front();
    int x2 = q.front();
    q.pop_front();
    if (x1 > x2) {
      q.push_front(x1);
      q.push_back(x2);
    } else {
      q.push_front(x2);
      q.push_back(x1);
    }
    tot++;
    v.push_back({x1, x2});
  }
  vector<int> vv;
  q.pop_front();
  while (q.size()) {
    vv.push_back(q.front());
    q.pop_front();
  }
  while (q1--) {
    long long x;
    cin >> x;
    if (x <= tot) {
      cout << v[x].first << " " << v[x].second << "\n";
    } else {
      cout << mx << " ";
      long long xx = (x - tot - 1) % (n - 1);
      cout << vv[xx] << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```