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;
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  out << '(' << p.first << ',' << p.second << ')';
  return out;
}
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <class T>
istream& operator>>(istream& in, vector<T>& v) {
  for (auto& x : v) in >> x;
  return in;
}
template <class T>
ostream& operator<<(ostream& out, vector<vector<T>>& v) {
  for (auto& x : v) out << x << '\n';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T>& v) {
  for (auto x : v) out << x << ' ';
  return out;
}
long long gcd(long long a, long long b) {
  if (b > a) swap(a, b);
  return (b ? gcd(b, a % b) : a);
}
const long double pi = 2 * asin(1);
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.setstate(ios::failbit);
  int n, q;
  cin >> n >> q;
  vector<long long> v(n);
  cin >> v;
  long long i = 0;
  int mx = *max_element(v.begin(), v.end());
  map<long long, pair<int, int>> m;
  while (1) {
    if (v[i] == mx) break;
    m[i + 1] = make_pair(v[i], v[i + 1]);
    if (v[i] > v[i + 1]) swap(v[i], v[i + 1]);
    v.push_back(v[i]);
    i++;
  }
  cerr << v << ' ' << i << '\n';
  while (q--) {
    long long x;
    cin >> x;
    if (m.count(x))
      cout << m[x].first << ' ' << m[x].second << '\n';
    else {
      x -= i + 1;
      cout << v[i] << ' ' << v[x % (n - 1) + i + 1] << '\n';
    }
  }
}
