#include <bits/stdc++.h>
using namespace std;
void fastIO(void);
void decimal(int);
long long int modmulti(long long int, long long int, long long int);
long long int modadd(long long int, long long int, long long int);
long long int modpower(long long int, long long int, long long int);
long long int gcd(long long int, long long int);
long long int modinv(long long int, long long int);
int sum(int, int);
bool check(long long int a[], long long int n) {
  for (long long int i = 1; i < n; i++) {
    if (a[i] < a[i - 1]) return 0;
  }
  return 1;
}
long long int nodigit(long long int n) {
  long long int count = 0;
  while (n != 0) {
    count++;
    n = n / 10;
  }
  return count;
}
long long int digit[11];
int main() {
  fastIO();
  long long int n, q;
  cin >> n >> q;
  deque<long long int> queue;
  long long int m;
  vector<pair<long long int, long long int> > ans;
  long long int mx = -1;
  for (long long int i = 0; i < n; i++) {
    cin >> m;
    queue.push_back(m);
    mx = max(mx, m);
  }
  long long int t1, t2;
  while (mx != queue.front()) {
    t1 = queue.front();
    queue.pop_front();
    t2 = queue.front();
    queue.pop_front();
    ans.push_back(make_pair(t1, t2));
    if (t1 > t2) {
      queue.push_back(t2);
      queue.push_front(t1);
    } else {
      queue.push_back(t1);
      queue.push_front(t2);
    }
  }
  for (long long int i = 0; i < q; i++) {
    cin >> m;
    if (m <= ans.size()) {
      cout << ans[m - 1].first << " " << ans[m - 1].second << "\n";
    } else {
      m = m - ans.size();
      m = m % (n - 1);
      if (m == 0) m = n - 1;
      cout << queue.front() << " " << queue.at(m) << "\n";
    }
  }
}
void fastIO(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
void decimal(int n) {
  cout << fixed;
  cout << setprecision(n);
}
long long int modmulti(long long int a, long long int b, long long int m) {
  return ((a % m) * (b % m)) % m;
}
long long int modadd(long long int a, long long int b, long long int m) {
  return ((a % m) + (b % m)) % m;
}
long long int modpower(long long int a, long long int n, long long int m) {
  if (n == 0) return 1;
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = modmulti(result, a, m);
    a = modmulti(a, a, m);
    n = n / 2;
  }
  return result;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int modinv(long long int a, long long int m) {
  return modpower(a, m - 2, m);
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
void fastIO(void);
void decimal(int);
long long int modmulti(long long int, long long int, long long int);
long long int modadd(long long int, long long int, long long int);
long long int modpower(long long int, long long int, long long int);
long long int gcd(long long int, long long int);
long long int modinv(long long int, long long int);
int sum(int, int);
bool check(long long int a[], long long int n) {
  for (long long int i = 1; i < n; i++) {
    if (a[i] < a[i - 1]) return 0;
  }
  return 1;
}
long long int nodigit(long long int n) {
  long long int count = 0;
  while (n != 0) {
    count++;
    n = n / 10;
  }
  return count;
}
long long int digit[11];
int main() {
  fastIO();
  long long int n, q;
  cin >> n >> q;
  deque<long long int> queue;
  long long int m;
  vector<pair<long long int, long long int> > ans;
  long long int mx = -1;
  for (long long int i = 0; i < n; i++) {
    cin >> m;
    queue.push_back(m);
    mx = max(mx, m);
  }
  long long int t1, t2;
  while (mx != queue.front()) {
    t1 = queue.front();
    queue.pop_front();
    t2 = queue.front();
    queue.pop_front();
    ans.push_back(make_pair(t1, t2));
    if (t1 > t2) {
      queue.push_back(t2);
      queue.push_front(t1);
    } else {
      queue.push_back(t1);
      queue.push_front(t2);
    }
  }
  for (long long int i = 0; i < q; i++) {
    cin >> m;
    if (m <= ans.size()) {
      cout << ans[m - 1].first << " " << ans[m - 1].second << "\n";
    } else {
      m = m - ans.size();
      m = m % (n - 1);
      if (m == 0) m = n - 1;
      cout << queue.front() << " " << queue.at(m) << "\n";
    }
  }
}
void fastIO(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
void decimal(int n) {
  cout << fixed;
  cout << setprecision(n);
}
long long int modmulti(long long int a, long long int b, long long int m) {
  return ((a % m) * (b % m)) % m;
}
long long int modadd(long long int a, long long int b, long long int m) {
  return ((a % m) + (b % m)) % m;
}
long long int modpower(long long int a, long long int n, long long int m) {
  if (n == 0) return 1;
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = modmulti(result, a, m);
    a = modmulti(a, a, m);
    n = n / 2;
  }
  return result;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int modinv(long long int a, long long int m) {
  return modpower(a, m - 2, m);
}

```