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;
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
long long power(long long B, long long P) {
  if (P == 0) return 1;
  long long X = power(B, P / 2);
  if (P % 2 == 0)
    return X * X;
  else
    return B * X * X;
}
long long fx4[] = {1, -1, 0, 0};
long long fy4[] = {0, 0, 1, -1};
long long A[400005];
int main() {
  long long T, N, M, X, Y, W, K, Q, R, P;
  cin >> N >> Q;
  deque<long long> V;
  long long maxv = 0, maxi = 1;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    V.push_back(A[i]);
    if (A[i] > maxv) {
      maxv = A[i];
      maxi = i;
    }
  }
  vector<pair<long long, long long> > ans;
  ans.push_back(make_pair(0, 0));
  while (1) {
    long long u1 = V.front();
    V.pop_front();
    long long u2 = V.front();
    V.pop_front();
    ans.push_back(make_pair(u1, u2));
    if (u1 > u2) {
      V.push_front(u1);
      V.push_back(u2);
    } else {
      V.push_front(u2);
      V.push_back(u1);
    }
    if (u1 == maxv) break;
  }
  for (int i = 1; i <= Q; i++) {
    long long op;
    cin >> op;
    if (op <= maxi) {
      cout << ans[op].first << " " << ans[op].second << endl;
    } else {
      long long x = op - maxi;
      long long y = x % (N - 1);
      if (y == 0) y = N - 1;
      cout << maxv << " " << V[y] << endl;
    }
  }
  return 0;
}
