#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 123, MAXN = 5e5 + 47, MEGAINF = 1e18 + 228;
template <class T>
inline istream& operator>>(istream& in, vector<T>& a) {
  for (auto& i : a) in >> i;
  return in;
}
template <class T>
inline ostream& operator<<(ostream& out, vector<T>& a) {
  for (auto i : a) out << i << " ";
  return out;
}
template <class T, class U>
inline istream& operator>>(istream& in, vector<pair<T, U>>& a) {
  for (auto& i : a) in >> i.first >> i.second;
  return in;
}
template <class T, class U>
inline ostream& operator<<(ostream& out, vector<pair<T, U>>& a) {
  for (auto& i : a) out << i.first << " " << i.second << "\n";
  return out;
}
signed main() {
  setlocale(LC_ALL, "rus");
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  vector<pair<long long, long long>> p;
  for (long long i = 0; i < n; ++i) p.push_back({a[i], i});
  sort(p.begin(), p.end(),
       [&](pair<long long, long long> one, pair<long long, long long> two) {
         if (one.first == two.first) return one.second < two.second;
         return one.first > two.first;
       });
  long long m;
  cin >> m;
  while (m--) {
    long long k, pos;
    cin >> k >> pos;
    --pos;
    vector<long long> have;
    for (long long i = 0; i < k; ++i) have.push_back(p[i].second);
    sort(have.begin(), have.end());
    long long ind = have[pos];
    cout << a[ind] << "\n";
  }
}

### Prompt

Your challenge is to write a CPP solution to 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;
const long long INF = 1e9 + 123, MAXN = 5e5 + 47, MEGAINF = 1e18 + 228;
template <class T>
inline istream& operator>>(istream& in, vector<T>& a) {
  for (auto& i : a) in >> i;
  return in;
}
template <class T>
inline ostream& operator<<(ostream& out, vector<T>& a) {
  for (auto i : a) out << i << " ";
  return out;
}
template <class T, class U>
inline istream& operator>>(istream& in, vector<pair<T, U>>& a) {
  for (auto& i : a) in >> i.first >> i.second;
  return in;
}
template <class T, class U>
inline ostream& operator<<(ostream& out, vector<pair<T, U>>& a) {
  for (auto& i : a) out << i.first << " " << i.second << "\n";
  return out;
}
signed main() {
  setlocale(LC_ALL, "rus");
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  vector<pair<long long, long long>> p;
  for (long long i = 0; i < n; ++i) p.push_back({a[i], i});
  sort(p.begin(), p.end(),
       [&](pair<long long, long long> one, pair<long long, long long> two) {
         if (one.first == two.first) return one.second < two.second;
         return one.first > two.first;
       });
  long long m;
  cin >> m;
  while (m--) {
    long long k, pos;
    cin >> k >> pos;
    --pos;
    vector<long long> have;
    for (long long i = 0; i < k; ++i) have.push_back(p[i].second);
    sort(have.begin(), have.end());
    long long ind = have[pos];
    cout << a[ind] << "\n";
  }
}

```