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;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
bool prime[10000001];
long long P = 1000000007 - 2;
long long factorial[10000001] = {0};
long long powerFunction(long long a, long long n) {
  long long r;
  if (n == 0)
    return 1;
  else if (n == 1)
    return a;
  else
    r = powerFunction(a, n / 2);
  if (n % 2 == 0)
    return ((r) % 1000000007) * ((r) % 1000000007) % 1000000007;
  else
    return ((r) % 1000000007) * ((r) % 1000000007) * ((a) % 1000000007) %
           1000000007;
}
long long modInverseFunctionFunction(long long no) {
  long long inverse = 1;
  while (P > 0) {
    if (P & 1) {
      inverse = (inverse % 1000000007 * no % 1000000007) % 1000000007;
    }
    no = (no % 1000000007 * no % 1000000007) % 1000000007;
    P = P / 2;
  }
  P = 1000000007 - 2;
  return inverse % 1000000007;
}
void sieveFunction(long long maxLimit) {
  memset(prime, true, sizeof(prime));
  prime[0] = prime[1] = false;
  for (long long i = 2; maxLimit >= i; i++) {
    if (prime[i]) {
      for (long long j = 2 * i; maxLimit > j; j += i) prime[j] = false;
    }
  }
}
void factorialFunction(long long maxLimit) {
  factorial[0] = 1;
  for (long long i = 1; i <= maxLimit; i++)
    factorial[i] = (factorial[i - 1] * i) % 1000000007;
  return;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, t;
  cin >> n >> t;
  long long arr[n];
  long long mx = INT_MIN;
  long long idx = -1;
  deque<long long> q;
  for (long long i = 0; n > i; i++) {
    cin >> arr[i];
    q.push_back(arr[i]);
    if (arr[i] >= mx) {
      mx = arr[i];
      idx = i;
    }
  }
  vector<long long> v;
  vector<pair<long long, long long>> vp;
  long long count = 0;
  while (count != idx) {
    long long top = q.front();
    q.pop_front();
    long long newTop = q.front();
    vp.push_back({top, newTop});
    if (newTop > top) {
      q.push_back(top);
    } else {
      q.pop_front();
      q.push_back(newTop);
      q.push_front(top);
    }
    count++;
  }
  q.pop_front();
  while (q.size()) {
    long long p = q.front();
    q.pop_front();
    v.push_back(p);
  }
  while (t--) {
    long long m;
    cin >> m;
    if (m <= idx) {
      cout << vp[m - 1].first << ' ' << vp[m - 1].second << "\n";
    } else {
      m -= idx;
      m--;
      m = (m + 2 * (n - 1)) % (n - 1);
      cout << mx << ' ' << v[m] << "\n";
    }
  }
}
