#include <bits/stdc++.h>
using namespace std;
vector<bool> sieve(long long n) {
  vector<bool> prime(n + 1, true);
  prime[0] = prime[1] = false;
  for (long long i = 2; i <= n; ++i) {
    if (prime[i] && (i * i) <= n)
      for (long long j = i * i; j <= n; j += i) prime[j] = false;
  }
  return prime;
}
long long power(long long a, long long b, long long m = 1000000007) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, q;
  cin >> n >> q;
  deque<long long> d;
  long long mx = 0;
  vector<pair<long long, long long> > res;
  for (long long i = 0; i < n; ++i) {
    long long a;
    cin >> a;
    d.push_back(a);
    mx = max(mx, a);
  }
  while (d.front() != mx) {
    long long a = d[0], b = d[1];
    d.pop_front();
    d.pop_front();
    d.push_front(max(a, b));
    d.push_back(min(a, b));
    res.push_back({a, b});
  }
  while (q--) {
    long long m;
    cin >> m;
    m -= 1;
    if (m < (long long)res.size())
      cout << res[m].first << " " << res[m].second << '\n';
    else
      cout << d[0] << " " << d[(m - (long long)res.size()) % (n - 1) + 1]
           << '\n';
  }
  return 0;
}

### Prompt

Generate 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;
vector<bool> sieve(long long n) {
  vector<bool> prime(n + 1, true);
  prime[0] = prime[1] = false;
  for (long long i = 2; i <= n; ++i) {
    if (prime[i] && (i * i) <= n)
      for (long long j = i * i; j <= n; j += i) prime[j] = false;
  }
  return prime;
}
long long power(long long a, long long b, long long m = 1000000007) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, q;
  cin >> n >> q;
  deque<long long> d;
  long long mx = 0;
  vector<pair<long long, long long> > res;
  for (long long i = 0; i < n; ++i) {
    long long a;
    cin >> a;
    d.push_back(a);
    mx = max(mx, a);
  }
  while (d.front() != mx) {
    long long a = d[0], b = d[1];
    d.pop_front();
    d.pop_front();
    d.push_front(max(a, b));
    d.push_back(min(a, b));
    res.push_back({a, b});
  }
  while (q--) {
    long long m;
    cin >> m;
    m -= 1;
    if (m < (long long)res.size())
      cout << res[m].first << " " << res[m].second << '\n';
    else
      cout << d[0] << " " << d[(m - (long long)res.size()) % (n - 1) + 1]
           << '\n';
  }
  return 0;
}

```