#include <bits/stdc++.h>
const long long INF = 1e9 + 7, mod = 998244353, MAXN = 200, SQ = 322, p = 31;
const long double EPS = 1e-6, PI = 3.141592653589793238462643;
using namespace std;
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("profile-values,profile-reorder-functions,tracer")
#pragma GCC optimize("vpt")
#pragma GCC optimize("rename-registers")
#pragma GCC optimize("move-loop-invariants")
#pragma GCC optimize("unswitch-loops")
#pragma GCC optimize("function-sections")
#pragma GCC optimize("data-sections")
#pragma GCC optimize("branch-target-load-optimize")
#pragma GCC optimize("btr-bb-exclusive")
namespace in_out {
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (auto i : a) out << i << " ";
  out << '\n';
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (long long i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
}  // namespace in_out
using namespace in_out;
vector<int> t;
void upd(int v, int l, int r, int i) {
  if (l >= r || l > i || r <= i) return;
  if (r - l == 1 && l == i) {
    t[v] += 1;
    return;
  }
  int m = (l + r) / 2;
  upd(2 * v, l, m, i);
  upd(2 * v + 1, m, r, i);
  t[v] = t[2 * v] + t[2 * v + 1];
}
int qry(int v, int l, int r, int x) {
  if (r - l == 1) return l;
  int m = (l + r) / 2;
  if (t[2 * v] >= x) return qry(2 * v, l, m, x);
  return qry(2 * v + 1, m, r, x - t[2 * v]);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  cin >> a;
  t.resize(4 * n);
  vector<pair<int, int>> b(n);
  for (int i = 0; i < n; i++) b[i] = make_pair(a[i], -i);
  sort(b.rbegin(), b.rend());
  vector<vector<pair<int, int>>> q(n + 1);
  int qq;
  cin >> qq;
  for (int i = 0; i < qq; i++) {
    int k, pos;
    cin >> k >> pos;
    q[k].push_back({pos, i});
  }
  vector<int> ans(qq);
  for (int i = 1; i < n + 1; i++) {
    upd(1, 0, n, -b[i - 1].second);
    if (q[i].size() == 0) continue;
    for (auto& pos : q[i]) {
      int ass = qry(1, 0, n, pos.first);
      ans[pos.second] = a[ass];
    }
  }
  for (auto& i : ans) cout << i << "\n";
  return 0;
}

### 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>
const long long INF = 1e9 + 7, mod = 998244353, MAXN = 200, SQ = 322, p = 31;
const long double EPS = 1e-6, PI = 3.141592653589793238462643;
using namespace std;
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("profile-values,profile-reorder-functions,tracer")
#pragma GCC optimize("vpt")
#pragma GCC optimize("rename-registers")
#pragma GCC optimize("move-loop-invariants")
#pragma GCC optimize("unswitch-loops")
#pragma GCC optimize("function-sections")
#pragma GCC optimize("data-sections")
#pragma GCC optimize("branch-target-load-optimize")
#pragma GCC optimize("btr-bb-exclusive")
namespace in_out {
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (auto i : a) out << i << " ";
  out << '\n';
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (long long i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
}  // namespace in_out
using namespace in_out;
vector<int> t;
void upd(int v, int l, int r, int i) {
  if (l >= r || l > i || r <= i) return;
  if (r - l == 1 && l == i) {
    t[v] += 1;
    return;
  }
  int m = (l + r) / 2;
  upd(2 * v, l, m, i);
  upd(2 * v + 1, m, r, i);
  t[v] = t[2 * v] + t[2 * v + 1];
}
int qry(int v, int l, int r, int x) {
  if (r - l == 1) return l;
  int m = (l + r) / 2;
  if (t[2 * v] >= x) return qry(2 * v, l, m, x);
  return qry(2 * v + 1, m, r, x - t[2 * v]);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  cin >> a;
  t.resize(4 * n);
  vector<pair<int, int>> b(n);
  for (int i = 0; i < n; i++) b[i] = make_pair(a[i], -i);
  sort(b.rbegin(), b.rend());
  vector<vector<pair<int, int>>> q(n + 1);
  int qq;
  cin >> qq;
  for (int i = 0; i < qq; i++) {
    int k, pos;
    cin >> k >> pos;
    q[k].push_back({pos, i});
  }
  vector<int> ans(qq);
  for (int i = 1; i < n + 1; i++) {
    upd(1, 0, n, -b[i - 1].second);
    if (q[i].size() == 0) continue;
    for (auto& pos : q[i]) {
      int ass = qry(1, 0, n, pos.first);
      ans[pos.second] = a[ass];
    }
  }
  for (auto& i : ans) cout << i << "\n";
  return 0;
}

```