#include <bits/stdc++.h>
using namespace std;
const double PI = 2 * acos(0.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());
const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";
template <class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
  return out << "{" << a.first << ", " << a.second << "}";
}
template <class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << "}";
}
template <class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << ", " << get<3>(a) << "}";
}
template <class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
  out << "[";
  for (int i = 0; i < int(a.size()); ++i)
    out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
  return out;
}
void smain();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(12) << fixed;
  smain();
  return 0;
}
const int N = 2e5 + 100;
int t[N << 2];
void upd(int v, int tl, int tr, int pos) {
  t[v]++;
  if (tl == tr) {
    return;
  }
  int tm = tl + tr >> 1;
  if (pos <= tm)
    upd(v << 1, tl, tm, pos);
  else
    upd(v << 1 | 1, tm + 1, tr, pos);
}
int go(int v, int tl, int tr, int bound) {
  if (tl == tr) return tl;
  int tm = tl + tr >> 1;
  if (t[v << 1] >= bound) return go(v << 1, tl, tm, bound);
  return go(v << 1 | 1, tm + 1, tr, bound - t[v << 1]);
}
int ans[N];
vector<pair<int, int> > query[N];
void smain() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i];
  }
  vector<pair<int, int> > b;
  for (int i = 0; i < int(n); ++i) {
    b.push_back({-a[i], i});
  }
  sort(b.begin(), b.end());
  int q;
  cin >> q;
  for (int i = 0; i < int(q); ++i) {
    int k, pos;
    cin >> k >> pos;
    query[k].push_back({i, pos});
  }
  for (int k = 1; k <= n; ++k) {
    upd(1, 0, N - 1, b[k - 1].second);
    for (auto item : query[k]) {
      int pos = item.second;
      int id = go(1, 0, N - 1, pos);
      ans[item.first] = a[id];
    }
  }
  for (int i = 0; i < q; ++i) {
    cout << ans[i] << '\n';
  }
}

### 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;
const double PI = 2 * acos(0.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());
const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";
template <class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
  return out << "{" << a.first << ", " << a.second << "}";
}
template <class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << "}";
}
template <class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << ", " << get<3>(a) << "}";
}
template <class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
  out << "[";
  for (int i = 0; i < int(a.size()); ++i)
    out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
  return out;
}
void smain();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(12) << fixed;
  smain();
  return 0;
}
const int N = 2e5 + 100;
int t[N << 2];
void upd(int v, int tl, int tr, int pos) {
  t[v]++;
  if (tl == tr) {
    return;
  }
  int tm = tl + tr >> 1;
  if (pos <= tm)
    upd(v << 1, tl, tm, pos);
  else
    upd(v << 1 | 1, tm + 1, tr, pos);
}
int go(int v, int tl, int tr, int bound) {
  if (tl == tr) return tl;
  int tm = tl + tr >> 1;
  if (t[v << 1] >= bound) return go(v << 1, tl, tm, bound);
  return go(v << 1 | 1, tm + 1, tr, bound - t[v << 1]);
}
int ans[N];
vector<pair<int, int> > query[N];
void smain() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i];
  }
  vector<pair<int, int> > b;
  for (int i = 0; i < int(n); ++i) {
    b.push_back({-a[i], i});
  }
  sort(b.begin(), b.end());
  int q;
  cin >> q;
  for (int i = 0; i < int(q); ++i) {
    int k, pos;
    cin >> k >> pos;
    query[k].push_back({i, pos});
  }
  for (int k = 1; k <= n; ++k) {
    upd(1, 0, N - 1, b[k - 1].second);
    for (auto item : query[k]) {
      int pos = item.second;
      int id = go(1, 0, N - 1, pos);
      ans[item.first] = a[id];
    }
  }
  for (int i = 0; i < q; ++i) {
    cout << ans[i] << '\n';
  }
}

```