#include <bits/stdc++.h>
using namespace std;
inline void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  map<int, vector<int> > mp;
  for (int i = 0; i < n; ++i) {
    mp[a[i]].push_back(i);
  }
  vector<int> szs(0);
  vector<int> ind(0);
  for (auto i : mp) {
    szs.push_back((int)i.second.size());
    ind.push_back((int)i.first);
  }
  reverse(szs.begin(), szs.end());
  reverse(ind.begin(), ind.end());
  vector<int> ps = {szs[0]};
  for (int i = 1; i < szs.size(); ++i) {
    ps.push_back(ps.back() + szs[i]);
  }
  int m;
  cin >> m;
  while (m--) {
    int k, pos;
    cin >> k >> pos;
    pos--;
    set<int> psps;
    int nowind = 0;
    int nowel = 0;
    for (int i = 0; i < k; ++i) {
      psps.insert(mp[ind[nowind]][nowel]);
      nowel++;
      if (nowel == mp[ind[nowind]].size()) {
        nowel = 0;
        nowind++;
      }
    }
    vector<int> v;
    for (auto i : psps) {
      v.push_back(i);
    }
    cout << a[v[pos]] << "\n";
  }
}
signed main() {
  fastio();
  solve();
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
inline void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  map<int, vector<int> > mp;
  for (int i = 0; i < n; ++i) {
    mp[a[i]].push_back(i);
  }
  vector<int> szs(0);
  vector<int> ind(0);
  for (auto i : mp) {
    szs.push_back((int)i.second.size());
    ind.push_back((int)i.first);
  }
  reverse(szs.begin(), szs.end());
  reverse(ind.begin(), ind.end());
  vector<int> ps = {szs[0]};
  for (int i = 1; i < szs.size(); ++i) {
    ps.push_back(ps.back() + szs[i]);
  }
  int m;
  cin >> m;
  while (m--) {
    int k, pos;
    cin >> k >> pos;
    pos--;
    set<int> psps;
    int nowind = 0;
    int nowel = 0;
    for (int i = 0; i < k; ++i) {
      psps.insert(mp[ind[nowind]][nowel]);
      nowel++;
      if (nowel == mp[ind[nowind]].size()) {
        nowel = 0;
        nowind++;
      }
    }
    vector<int> v;
    for (auto i : psps) {
      v.push_back(i);
    }
    cout << a[v[pos]] << "\n";
  }
}
signed main() {
  fastio();
  solve();
}

```