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. 
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
int mod = 1e9 + 7;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long powmod(long long a, long long b) {
  long long res = 1;
  if (a >= mod) a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    if (res >= mod) res %= mod;
    a = a * a;
    if (a >= mod) a %= mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int a[300005];
  int res[300005];
  pair<int, int> ans[300005];
  int n, q;
  cin >> n >> q;
  int TOP = -1, idx = -1;
  for (long long i = (long long)(1); i < (long long)(n + 1); i++) cin >> a[i];
  for (long long i = (long long)(1); i < (long long)(n + 1); i++) {
    if (a[i] > TOP) {
      TOP = a[i];
      idx = i;
    }
  }
  long long z = n;
  for (long long i = (long long)(2); i < (long long)(idx + 1); i++) {
    ans[i - 1] = make_pair(a[1], a[i]);
    if (a[1] > a[i]) {
      a[++z] = a[i];
    } else {
      swap(a[1], a[i]);
      a[++z] = a[i];
    }
  }
  a[idx] = TOP;
  int j = 0;
  for (long long i = (long long)(idx); i < (long long)(z + 1); i++)
    res[++j] = a[i];
  while (q--) {
    long long x;
    cin >> x;
    if (x >= idx) {
      x -= (idx - 1);
      x %= (n - 1);
      cout << TOP << " ";
      if (x == 0)
        cout << res[n] << '\n';
      else
        cout << res[x + 1] << '\n';
    } else {
      cout << ans[x].first << " " << ans[x].second << '\n';
    }
  }
  return 0;
}
