#include <bits/stdc++.h>
using namespace std;
void itval(istream_iterator<string> it) {}
template <typename T, typename... Args>
void itval(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  itval(++it, args...);
}
template <typename T>
inline void print(T x) {
  cout << x << "\n";
}
template <typename T>
inline void printvec(T x) {
  for (auto a : x) cout << a << ' ';
  cout << '\n';
}
const long long int MOD = 1e9 + 7;
struct custom {
  bool operator()(const pair<int, int> &p1, const pair<int, int> &p2) const {
    if (p1.first == p2.first) {
      return p1.second < p2.second;
    }
    return p1.first > p2.first;
  }
};
long long int get_pow(long long int x, long long int k) {
  if (k == 0) return 1ll;
  long long int y = get_pow(x, k / 2);
  y = (y * y) % MOD;
  if (k % 2) y = (y * x) % MOD;
  return y;
}
const int N = 2e5 + 10;
void solve() {
  int n, m, k, x;
  cin >> n;
  vector<int> v(n);
  for (int i = (int)0; i < int(n); i++) cin >> v[i];
  vector<pair<int, int> > res;
  for (int i = (int)0; i < int(n); i++) {
    res.push_back({v[i], i});
  }
  sort(res.begin(), res.end(), custom());
  cin >> m;
  for (int i = (int)0; i < int(m); i++) {
    cin >> k >> x;
    x--;
    vector<int> dum;
    for (int i = (int)0; i < int(k); i++) dum.push_back(res[i].second);
    sort(dum.begin(), dum.end());
    for (int i = (int)0; i < int(k); i++) dum[i] = v[dum[i]];
    cout << dum[x] << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  clock_t z = clock();
  for (int tes = (int)0; tes < int(test); tes++) {
    solve();
  }
  fprintf(stderr, "Total Time:%.4f\n", (double)(clock() - z) / CLOCKS_PER_SEC),
      fflush(stderr);
  return 0;
}

### Prompt

Generate 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;
void itval(istream_iterator<string> it) {}
template <typename T, typename... Args>
void itval(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  itval(++it, args...);
}
template <typename T>
inline void print(T x) {
  cout << x << "\n";
}
template <typename T>
inline void printvec(T x) {
  for (auto a : x) cout << a << ' ';
  cout << '\n';
}
const long long int MOD = 1e9 + 7;
struct custom {
  bool operator()(const pair<int, int> &p1, const pair<int, int> &p2) const {
    if (p1.first == p2.first) {
      return p1.second < p2.second;
    }
    return p1.first > p2.first;
  }
};
long long int get_pow(long long int x, long long int k) {
  if (k == 0) return 1ll;
  long long int y = get_pow(x, k / 2);
  y = (y * y) % MOD;
  if (k % 2) y = (y * x) % MOD;
  return y;
}
const int N = 2e5 + 10;
void solve() {
  int n, m, k, x;
  cin >> n;
  vector<int> v(n);
  for (int i = (int)0; i < int(n); i++) cin >> v[i];
  vector<pair<int, int> > res;
  for (int i = (int)0; i < int(n); i++) {
    res.push_back({v[i], i});
  }
  sort(res.begin(), res.end(), custom());
  cin >> m;
  for (int i = (int)0; i < int(m); i++) {
    cin >> k >> x;
    x--;
    vector<int> dum;
    for (int i = (int)0; i < int(k); i++) dum.push_back(res[i].second);
    sort(dum.begin(), dum.end());
    for (int i = (int)0; i < int(k); i++) dum[i] = v[dum[i]];
    cout << dum[x] << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  clock_t z = clock();
  for (int tes = (int)0; tes < int(test); tes++) {
    solve();
  }
  fprintf(stderr, "Total Time:%.4f\n", (double)(clock() - z) / CLOCKS_PER_SEC),
      fflush(stderr);
  return 0;
}

```