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;
inline void prnt(bool ok) { cout << (ok ? "YES" : "NO"); }
const int ppr = 257;
const long long INF = 2e18;
const int inf = 2e9;
const int mod = 1e9 + 9;
const int N = 1e6 + 123;
const long double pi = 3.141592653589793238462643;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
long long n, a[N], b[N], q, m, mx = -1, pos;
deque<long long> d;
pair<int, int> p[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    d.push_back(a[i]);
    mx = max(mx, a[i]);
  }
  for (long long i = 1; i <= n; i++) {
    if (mx == a[i]) {
      pos = i;
      break;
    }
  }
  long long k = 0;
  while (true) {
    if (d.front() == mx) break;
    long long a = d.front();
    d.pop_front();
    long long b = d.front();
    d.pop_front();
    if (a > b)
      d.push_front(a), d.push_back(b);
    else
      d.push_front(b), d.push_back(a);
    p[++k] = make_pair(a, b);
  }
  k = 0;
  while (!d.empty()) {
    b[++k] = d.front();
    d.pop_front();
  }
  pos--;
  while (q--) {
    cin >> m;
    if (m <= pos) {
      cout << p[m].first << ' ' << p[m].second << '\n';
    } else {
      m -= pos;
      m--;
      m %= (n - 1);
      cout << b[1] << ' ' << b[1 + m + 1] << '\n';
    }
  }
  return 0;
}
