#include <bits/stdc++.h>
using namespace std;
mt19937 rng(static_cast<unsigned int>(
    chrono::steady_clock::now().time_since_epoch().count()));
template <typename num_t>
inline void add_mod(num_t& a, const long long& b, const num_t& m) {
  a = (a + b) % m;
  if (a < 0) a += m;
}
template <typename num_t>
inline bool update_max(num_t& a, const num_t& b) {
  return a < b ? a = b, true : false;
}
template <typename num_t>
inline bool update_min(num_t& a, const num_t& b) {
  return a > b ? a = b, true : false;
}
template <typename num_t>
num_t gcd(const num_t& lhs, const num_t& rhs) {
  return !lhs ? rhs : gcd(rhs % lhs, lhs);
}
template <typename num_t>
num_t lcm(const num_t& lhs, const num_t& rhs) {
  return lhs / ::gcd(lhs, rhs) * rhs;
}
template <typename num_t>
num_t pw(num_t n, num_t k, const num_t& mod) {
  if (k == -1) k = mod - 2;
  num_t res = 1;
  for (; k > 0; k >>= 1) {
    if (k & 1) res = 1ll * res * n % mod;
    n = 1ll * n * n % mod;
  }
  return res % mod;
}
const int inf = 1e9 + 7;
const int mod = 998244353;
const long long ll_inf = (long long)inf * inf * 9 + 7;
const int maxn = 150000 + 7;
template <typename num_t, typename merge = plus<num_t>,
          typename split = minus<num_t>>
struct FenWickTree {
  vector<num_t> tree;
  void update(int index, const num_t& value) {
    for (int t = index, max_t = (int)tree.size(); t < max_t; t += t & -t) {
      tree[t] = merge()(tree[t], value);
    }
  }
  num_t get(int index) {
    num_t res = 0;
    for (int t = index; t > 0; t -= t & -t) {
      res = merge()(res, tree[t]);
    }
    return res;
  }
  num_t get(int lhs, int rhs) { return split()(get(rhs), get(lhs - 1)); }
  void init(int n) { tree.resize(n + 1); }
};
FenWickTree<int> fenwick_tree;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto& x : a) cin >> x;
  vector<int> id(n);
  iota(id.begin(), id.end(), 0);
  sort(id.begin(), id.end(), [&](int u, int v) -> bool {
    return make_pair(a[u], -u) > make_pair(a[v], -v);
  });
  int m;
  cin >> m;
  vector<vector<pair<int, int>>> query(n + 1);
  for (int i = (0), _b = (m); i < _b; ++i) {
    int k, pos;
    cin >> k >> pos;
    query[k].push_back({i, pos});
  }
  vector<int> ans(m, 0);
  fenwick_tree.init(n + 2);
  for (int k = 1; k <= n; ++k) {
    int cur_pos = id[k - 1] + 1;
    fenwick_tree.update(cur_pos, 1);
    for (auto p : query[k]) {
      int idx = p.first;
      int pos = p.second;
      int start_index = 0;
      for (int j = 19; j >= 0; --j) {
        int nxt = start_index + (1 << j);
        if (nxt < n && fenwick_tree.get(nxt) < pos) start_index = nxt;
      }
      ans[idx] = a[start_index];
    }
  }
  for (auto x : ans) cout << x << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}

### Prompt

Please formulate 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(static_cast<unsigned int>(
    chrono::steady_clock::now().time_since_epoch().count()));
template <typename num_t>
inline void add_mod(num_t& a, const long long& b, const num_t& m) {
  a = (a + b) % m;
  if (a < 0) a += m;
}
template <typename num_t>
inline bool update_max(num_t& a, const num_t& b) {
  return a < b ? a = b, true : false;
}
template <typename num_t>
inline bool update_min(num_t& a, const num_t& b) {
  return a > b ? a = b, true : false;
}
template <typename num_t>
num_t gcd(const num_t& lhs, const num_t& rhs) {
  return !lhs ? rhs : gcd(rhs % lhs, lhs);
}
template <typename num_t>
num_t lcm(const num_t& lhs, const num_t& rhs) {
  return lhs / ::gcd(lhs, rhs) * rhs;
}
template <typename num_t>
num_t pw(num_t n, num_t k, const num_t& mod) {
  if (k == -1) k = mod - 2;
  num_t res = 1;
  for (; k > 0; k >>= 1) {
    if (k & 1) res = 1ll * res * n % mod;
    n = 1ll * n * n % mod;
  }
  return res % mod;
}
const int inf = 1e9 + 7;
const int mod = 998244353;
const long long ll_inf = (long long)inf * inf * 9 + 7;
const int maxn = 150000 + 7;
template <typename num_t, typename merge = plus<num_t>,
          typename split = minus<num_t>>
struct FenWickTree {
  vector<num_t> tree;
  void update(int index, const num_t& value) {
    for (int t = index, max_t = (int)tree.size(); t < max_t; t += t & -t) {
      tree[t] = merge()(tree[t], value);
    }
  }
  num_t get(int index) {
    num_t res = 0;
    for (int t = index; t > 0; t -= t & -t) {
      res = merge()(res, tree[t]);
    }
    return res;
  }
  num_t get(int lhs, int rhs) { return split()(get(rhs), get(lhs - 1)); }
  void init(int n) { tree.resize(n + 1); }
};
FenWickTree<int> fenwick_tree;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto& x : a) cin >> x;
  vector<int> id(n);
  iota(id.begin(), id.end(), 0);
  sort(id.begin(), id.end(), [&](int u, int v) -> bool {
    return make_pair(a[u], -u) > make_pair(a[v], -v);
  });
  int m;
  cin >> m;
  vector<vector<pair<int, int>>> query(n + 1);
  for (int i = (0), _b = (m); i < _b; ++i) {
    int k, pos;
    cin >> k >> pos;
    query[k].push_back({i, pos});
  }
  vector<int> ans(m, 0);
  fenwick_tree.init(n + 2);
  for (int k = 1; k <= n; ++k) {
    int cur_pos = id[k - 1] + 1;
    fenwick_tree.update(cur_pos, 1);
    for (auto p : query[k]) {
      int idx = p.first;
      int pos = p.second;
      int start_index = 0;
      for (int j = 19; j >= 0; --j) {
        int nxt = start_index + (1 << j);
        if (nxt < n && fenwick_tree.get(nxt) < pos) start_index = nxt;
      }
      ans[idx] = a[start_index];
    }
  }
  for (auto x : ans) cout << x << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}

```