#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;
  }
};
long long int gcd(long long int a, long long int b) {
  if (a == 0) {
    return b;
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long int power(long long int x, long long int n) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = (ans * x) % 1000000007;
    x = (x * x) % 1000000007;
    n /= 2;
  }
  if (x == 0) ans = 0;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    long long int n, i;
    cin >> n;
    vector<pair<long long int, long long int> > a;
    long long int pre[n];
    for (i = 0; i < n; i++) {
      cin >> pre[i];
      a.push_back(make_pair(pre[i], -1 * i));
    }
    sort(a.begin(), a.end());
    for (i = 1; i < n; i++) {
      a[i].first += (a[i - 1].first);
    }
    long long int m;
    cin >> m;
    for (i = 0; i < m; i++) {
      long long int k, r;
      cin >> k >> r;
      long long int maxs = INT_MIN, l;
      for (int j = n - 1; j >= k - 1; j--) {
        if (j != k - 1) {
          if (a[j].first - a[j - k].first >= maxs) {
            maxs = a[j].first - a[j - k].first;
            l = j - k + 1;
          }
        } else {
          if (a[j].first >= maxs) {
            maxs = a[j].first;
            l = 0;
          }
        }
      }
      vector<pair<long long int, long long int> > tmp;
      if (l == 0) {
        tmp.push_back(make_pair(-1 * a[0].second, a[0].first));
      }
      for (int z = l; z < l + k; z++) {
        long long int x = a[z].first, y = -1 * a[z].second;
        if (z) {
          tmp.push_back(make_pair(y, x - a[z - 1].first));
        }
      }
      sort(tmp.begin(), tmp.end());
      cout << tmp[r - 1].second << endl;
    }
  }
}

### Prompt

Create a solution in Cpp for the following problem:

This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.

You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:

  * [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list); 
  * [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences. 



Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:

  * it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k; 
  * and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal. 



Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:

  * [10, 20, 20] lexicographically less than [10, 21, 1], 
  * [7, 99, 99] is lexicographically less than [10, 21, 1], 
  * [10, 21, 0] is lexicographically less than [10, 21, 1]. 



You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.

For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.

The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).

The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.

The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.

Output

Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.

Examples

Input


3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3


Output


20
10
20
10
20
10


Input


7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4


Output


2
3
2
3
2
3
1
1
3

Note

In the first example, for a=[10,20,10] the optimal subsequences are: 

  * for k=1: [20], 
  * for k=2: [10,20], 
  * for k=3: [10,20,10]. 

### Response

```cpp
#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;
  }
};
long long int gcd(long long int a, long long int b) {
  if (a == 0) {
    return b;
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long int power(long long int x, long long int n) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = (ans * x) % 1000000007;
    x = (x * x) % 1000000007;
    n /= 2;
  }
  if (x == 0) ans = 0;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    long long int n, i;
    cin >> n;
    vector<pair<long long int, long long int> > a;
    long long int pre[n];
    for (i = 0; i < n; i++) {
      cin >> pre[i];
      a.push_back(make_pair(pre[i], -1 * i));
    }
    sort(a.begin(), a.end());
    for (i = 1; i < n; i++) {
      a[i].first += (a[i - 1].first);
    }
    long long int m;
    cin >> m;
    for (i = 0; i < m; i++) {
      long long int k, r;
      cin >> k >> r;
      long long int maxs = INT_MIN, l;
      for (int j = n - 1; j >= k - 1; j--) {
        if (j != k - 1) {
          if (a[j].first - a[j - k].first >= maxs) {
            maxs = a[j].first - a[j - k].first;
            l = j - k + 1;
          }
        } else {
          if (a[j].first >= maxs) {
            maxs = a[j].first;
            l = 0;
          }
        }
      }
      vector<pair<long long int, long long int> > tmp;
      if (l == 0) {
        tmp.push_back(make_pair(-1 * a[0].second, a[0].first));
      }
      for (int z = l; z < l + k; z++) {
        long long int x = a[z].first, y = -1 * a[z].second;
        if (z) {
          tmp.push_back(make_pair(y, x - a[z - 1].first));
        }
      }
      sort(tmp.begin(), tmp.end());
      cout << tmp[r - 1].second << endl;
    }
  }
}

```