#include <bits/stdc++.h>
using namespace std;
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
long long Sqrt(long long x) {
  if (x == 0 || x == 1) return x;
  long long start = 1, end = x, ans;
  while (start <= end) {
    long long mid = (start + end) / 2;
    if (mid * mid == x) return mid;
    if (mid * mid < x) {
      start = mid + 1;
      ans = mid;
    } else
      end = mid - 1;
  }
  return ans;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void in(long long &no) {
  bool neg = false;
  register long long c;
  no = 0;
  c = getchar();
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) no = no * 10 + c - 48;
  if (neg) no *= -1;
}
int main() {
  int n, i, t;
  long long z;
  cin >> n >> t;
  int a[n];
  for (long long i = (0); i < (n); i++) {
    cin >> a[i];
  }
  deque<int> dq;
  int max1 = a[0], itr = 0;
  dq.push_back(a[0]);
  for (long long i = (1); i < (n); i++) {
    dq.push_back(a[i]);
    if (a[i] > max1) {
      max1 = a[i];
      itr = i;
    }
  }
  pair<int, int> p1[itr + 1];
  int x, y;
  for (long long i = (0); i < (itr + 1); i++) {
    x = dq.front();
    dq.pop_front();
    y = dq.front();
    dq.pop_front();
    p1[i] = make_pair(x, y);
    dq.push_front(max(x, y));
    dq.push_back(min(x, y));
  }
  pair<int, int> p2[n];
  for (long long i = (0); i < (n); i++) {
    x = dq.front();
    dq.pop_front();
    y = dq.front();
    dq.pop_front();
    p2[i] = make_pair(x, y);
    dq.push_front(max(x, y));
    dq.push_back(min(x, y));
  }
  while (t--) {
    cin >> z;
    z--;
    if (z < itr + 1)
      cout << p1[z].first << " " << p1[z].second;
    else {
      z = z - itr - 1;
      z = z % (n - 1);
      cout << p2[z].first << " " << p2[z].second;
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
long long Sqrt(long long x) {
  if (x == 0 || x == 1) return x;
  long long start = 1, end = x, ans;
  while (start <= end) {
    long long mid = (start + end) / 2;
    if (mid * mid == x) return mid;
    if (mid * mid < x) {
      start = mid + 1;
      ans = mid;
    } else
      end = mid - 1;
  }
  return ans;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void in(long long &no) {
  bool neg = false;
  register long long c;
  no = 0;
  c = getchar();
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) no = no * 10 + c - 48;
  if (neg) no *= -1;
}
int main() {
  int n, i, t;
  long long z;
  cin >> n >> t;
  int a[n];
  for (long long i = (0); i < (n); i++) {
    cin >> a[i];
  }
  deque<int> dq;
  int max1 = a[0], itr = 0;
  dq.push_back(a[0]);
  for (long long i = (1); i < (n); i++) {
    dq.push_back(a[i]);
    if (a[i] > max1) {
      max1 = a[i];
      itr = i;
    }
  }
  pair<int, int> p1[itr + 1];
  int x, y;
  for (long long i = (0); i < (itr + 1); i++) {
    x = dq.front();
    dq.pop_front();
    y = dq.front();
    dq.pop_front();
    p1[i] = make_pair(x, y);
    dq.push_front(max(x, y));
    dq.push_back(min(x, y));
  }
  pair<int, int> p2[n];
  for (long long i = (0); i < (n); i++) {
    x = dq.front();
    dq.pop_front();
    y = dq.front();
    dq.pop_front();
    p2[i] = make_pair(x, y);
    dq.push_front(max(x, y));
    dq.push_back(min(x, y));
  }
  while (t--) {
    cin >> z;
    z--;
    if (z < itr + 1)
      cout << p1[z].first << " " << p1[z].second;
    else {
      z = z - itr - 1;
      z = z % (n - 1);
      cout << p2[z].first << " " << p2[z].second;
    }
    cout << "\n";
  }
  return 0;
}

```