#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000002;
const int MOD2 = 998244353;
const int MOD = 1000000007;
const int INF = 1e8;
const long double EPS = 1e-7;
long long int mul(long long int x, long long int y) {
  return (x * 1ll * y) % MOD2;
}
long long int fastpow(long long int x, long long int y) {
  long long int z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
long long int modinv(long long int n, long long int p) {
  return fastpow(n, p - 2);
}
struct Comp {
  bool operator()(const std::pair<int, int>& a, const std::pair<int, int>& b) {
    if (a.first != b.first) {
      return a.first > b.first;
    }
    return a.second > b.second;
  }
};
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  {
    long long int n, q1, t = 0;
    cin >> n >> q1;
    long long int a[n];
    deque<long long int> q;
    for (int i = 0; i < (int)n; i++) {
      cin >> a[i];
      q.push_back(a[i]);
    }
    long long int k = *max_element(a, a + n);
    vector<pair<long long int, long long int>> ans1, ans2;
    while (1) {
      long long int x = q.front();
      q.pop_front();
      long long int y = q.front();
      q.pop_front();
      long long int u = max(x, y);
      long long int v = min(x, y);
      ans1.push_back({x, y});
      t++;
      q.push_back(v);
      q.push_front(u);
      if (u == k) break;
    }
    q.pop_front();
    while (!q.empty()) {
      ans2.push_back({k, q.front()});
      q.pop_front();
    }
    while (q1--) {
      long long int p;
      cin >> p;
      if (p <= ans1.size())
        cout << (ans1[p - 1].first) << " " << (ans1[p - 1].second) << "\n";
      else {
        cout << max(ans2[(p - ans1.size() - 1) % (ans2.size())].first,
                    ans2[(p - ans1.size() - 1) % (ans2.size())].second)
             << " "
             << min(ans2[(p - ans1.size() - 1) % (ans2.size())].second,
                    ans2[(p - ans1.size() - 1) % (ans2.size())].first)
             << "\n";
      }
    }
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000002;
const int MOD2 = 998244353;
const int MOD = 1000000007;
const int INF = 1e8;
const long double EPS = 1e-7;
long long int mul(long long int x, long long int y) {
  return (x * 1ll * y) % MOD2;
}
long long int fastpow(long long int x, long long int y) {
  long long int z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
long long int modinv(long long int n, long long int p) {
  return fastpow(n, p - 2);
}
struct Comp {
  bool operator()(const std::pair<int, int>& a, const std::pair<int, int>& b) {
    if (a.first != b.first) {
      return a.first > b.first;
    }
    return a.second > b.second;
  }
};
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  {
    long long int n, q1, t = 0;
    cin >> n >> q1;
    long long int a[n];
    deque<long long int> q;
    for (int i = 0; i < (int)n; i++) {
      cin >> a[i];
      q.push_back(a[i]);
    }
    long long int k = *max_element(a, a + n);
    vector<pair<long long int, long long int>> ans1, ans2;
    while (1) {
      long long int x = q.front();
      q.pop_front();
      long long int y = q.front();
      q.pop_front();
      long long int u = max(x, y);
      long long int v = min(x, y);
      ans1.push_back({x, y});
      t++;
      q.push_back(v);
      q.push_front(u);
      if (u == k) break;
    }
    q.pop_front();
    while (!q.empty()) {
      ans2.push_back({k, q.front()});
      q.pop_front();
    }
    while (q1--) {
      long long int p;
      cin >> p;
      if (p <= ans1.size())
        cout << (ans1[p - 1].first) << " " << (ans1[p - 1].second) << "\n";
      else {
        cout << max(ans2[(p - ans1.size() - 1) % (ans2.size())].first,
                    ans2[(p - ans1.size() - 1) % (ans2.size())].second)
             << " "
             << min(ans2[(p - ans1.size() - 1) % (ans2.size())].second,
                    ans2[(p - ans1.size() - 1) % (ans2.size())].first)
             << "\n";
      }
    }
  }
  return 0;
}

```