#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void DEBUG(string label, T value) {
  cerr << "[" << label << " = " << value << "]\n";
}
class SegmentTree {
  const int n = 2e5 + 10;
  vector<int> st;

 public:
  SegmentTree() { st.resize(4 * n); }
  void update(int v, int tl, int tr, int pos) {
    if (tl == tr) {
      st[v] = 1;
    } else {
      int tm = (tl + tr) / 2;
      if (pos <= tm) {
        update(2 * v, tl, tm, pos);
      } else {
        update(2 * v + 1, tm + 1, tr, pos);
      }
      st[v] = st[2 * v] + st[2 * v + 1];
    }
  }
  int find_kth(int v, int tl, int tr, int k) {
    if (k > st[v]) {
      return -1;
    }
    if (tl == tr) {
      return tl;
    }
    int tm = (tl + tr) / 2;
    if (k <= st[2 * v]) {
      return find_kth(2 * v, tl, tm, k);
    } else {
      return find_kth(2 * v + 1, tm + 1, tr, k - st[2 * v]);
    }
  }
};
void solve() {
  int n, q;
  cin >> n;
  vector<pair<int, int>> arr(n);
  vector<int> original_array(n);
  for (int i = 0; i < n; ++i) {
    cin >> arr[i].first;
    arr[i].second = i;
    original_array[i] = arr[i].first;
  }
  sort(arr.begin(), arr.end(),
       [](const pair<int, int> &lhs, const pair<int, int> &rhs) -> bool {
         if (lhs.first != rhs.first) return lhs.first > rhs.first;
         return lhs.second < rhs.second;
       });
  cin >> q;
  vector<pair<pair<int, int>, int>> queries(q);
  for (int i = 0; i < q; ++i) {
    int k, p;
    cin >> k >> p;
    queries[i] = {{k, p}, i};
  }
  sort(queries.begin(), queries.end());
  int last = 0;
  vector<int> ans(q);
  SegmentTree st;
  const int N = 2e5 + 10;
  for (int i = 0; i < q; ++i) {
    int k = queries[i].first.first;
    int p = queries[i].first.second;
    int idx = queries[i].second;
    while (last < k) {
      st.update(1, 0, N, arr[last].second);
      last++;
    }
    int pos = st.find_kth(1, 0, N, p);
    ans[idx] = original_array[pos];
  }
  for (auto i : ans) {
    cout << i << '\n';
  }
}
int main(int argv, char **argc) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

### 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void DEBUG(string label, T value) {
  cerr << "[" << label << " = " << value << "]\n";
}
class SegmentTree {
  const int n = 2e5 + 10;
  vector<int> st;

 public:
  SegmentTree() { st.resize(4 * n); }
  void update(int v, int tl, int tr, int pos) {
    if (tl == tr) {
      st[v] = 1;
    } else {
      int tm = (tl + tr) / 2;
      if (pos <= tm) {
        update(2 * v, tl, tm, pos);
      } else {
        update(2 * v + 1, tm + 1, tr, pos);
      }
      st[v] = st[2 * v] + st[2 * v + 1];
    }
  }
  int find_kth(int v, int tl, int tr, int k) {
    if (k > st[v]) {
      return -1;
    }
    if (tl == tr) {
      return tl;
    }
    int tm = (tl + tr) / 2;
    if (k <= st[2 * v]) {
      return find_kth(2 * v, tl, tm, k);
    } else {
      return find_kth(2 * v + 1, tm + 1, tr, k - st[2 * v]);
    }
  }
};
void solve() {
  int n, q;
  cin >> n;
  vector<pair<int, int>> arr(n);
  vector<int> original_array(n);
  for (int i = 0; i < n; ++i) {
    cin >> arr[i].first;
    arr[i].second = i;
    original_array[i] = arr[i].first;
  }
  sort(arr.begin(), arr.end(),
       [](const pair<int, int> &lhs, const pair<int, int> &rhs) -> bool {
         if (lhs.first != rhs.first) return lhs.first > rhs.first;
         return lhs.second < rhs.second;
       });
  cin >> q;
  vector<pair<pair<int, int>, int>> queries(q);
  for (int i = 0; i < q; ++i) {
    int k, p;
    cin >> k >> p;
    queries[i] = {{k, p}, i};
  }
  sort(queries.begin(), queries.end());
  int last = 0;
  vector<int> ans(q);
  SegmentTree st;
  const int N = 2e5 + 10;
  for (int i = 0; i < q; ++i) {
    int k = queries[i].first.first;
    int p = queries[i].first.second;
    int idx = queries[i].second;
    while (last < k) {
      st.update(1, 0, N, arr[last].second);
      last++;
    }
    int pos = st.find_kth(1, 0, N, p);
    ans[idx] = original_array[pos];
  }
  for (auto i : ans) {
    cout << i << '\n';
  }
}
int main(int argv, char **argc) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

```