#include <bits/stdc++.h>
using namespace std;
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long int nxtl() {
  long long int x;
  cin >> x;
  return x;
}
void SieveOfEratosthenes(int n, vector<int> &primes) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) primes.push_back(p);
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int power(long long int x, long long int y, long long int mod) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2, mod);
  if (y % 2 == 0)
    return (temp * temp) % mod;
  else
    return (((x * temp) % mod) * temp) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n = nxt(), q = nxt();
  deque<int> q2;
  for (int i = 0; i < n; i++) {
    int x = nxt();
    q2.push_back(x);
  }
  vector<pair<int, int> > v(n + 1);
  for (int i = 1; i <= n; i++) {
    int x = q2.front();
    q2.pop_front();
    int y = q2.front();
    q2.pop_front();
    v[i] = {x, y};
    q2.push_back(min(x, y));
    q2.push_front(max(x, y));
  }
  vector<int> ans(n - 1);
  int x = q2.front();
  q2.pop_front();
  for (int i = 0; i < n - 1; i++) {
    ans[i] = q2.front();
    q2.pop_front();
    q2.push_back(ans[i]);
  }
  q2.push_front(x);
  while (q--) {
    long long int m = nxtl();
    long long int diff = m - n;
    if (diff <= 0) {
      cout << v[m].first << " " << v[m].second << endl;
      continue;
    }
    diff--;
    diff %= n - 1;
    cout << x << " " << ans[diff] << endl;
  }
  return 0;
}

### 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 nxt() {
  int x;
  cin >> x;
  return x;
}
long long int nxtl() {
  long long int x;
  cin >> x;
  return x;
}
void SieveOfEratosthenes(int n, vector<int> &primes) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) primes.push_back(p);
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int power(long long int x, long long int y, long long int mod) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2, mod);
  if (y % 2 == 0)
    return (temp * temp) % mod;
  else
    return (((x * temp) % mod) * temp) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n = nxt(), q = nxt();
  deque<int> q2;
  for (int i = 0; i < n; i++) {
    int x = nxt();
    q2.push_back(x);
  }
  vector<pair<int, int> > v(n + 1);
  for (int i = 1; i <= n; i++) {
    int x = q2.front();
    q2.pop_front();
    int y = q2.front();
    q2.pop_front();
    v[i] = {x, y};
    q2.push_back(min(x, y));
    q2.push_front(max(x, y));
  }
  vector<int> ans(n - 1);
  int x = q2.front();
  q2.pop_front();
  for (int i = 0; i < n - 1; i++) {
    ans[i] = q2.front();
    q2.pop_front();
    q2.push_back(ans[i]);
  }
  q2.push_front(x);
  while (q--) {
    long long int m = nxtl();
    long long int diff = m - n;
    if (diff <= 0) {
      cout << v[m].first << " " << v[m].second << endl;
      continue;
    }
    diff--;
    diff %= n - 1;
    cout << x << " " << ans[diff] << endl;
  }
  return 0;
}

```