#include <bits/stdc++.h>
using namespace std;
template <class T>
void dbs(string str, T t) {
  cout << str << " : " << t << "\n";
}
template <class T, class... second>
void dbs(string str, T t, second... s) {
  long long idx = str.find(',');
  cout << str.substr(0, idx) << " : " << t << ", ";
  dbs(str.substr(idx + 1), s...);
}
template <class T>
void prc(T a, T b) {
  cout << "[";
  for (T i = a; i != b; ++i) {
    if (i != a) cout << ", ";
    cout << *i;
  }
  cout << "]\n";
}
template <class T>
void prall(T a) {
  prc(a.begin(), a.end());
}
const int inf = 1000 * 1000 * 1000 + 9;
const long long infl = 1ll * inf * inf;
const int mod = 1000 * 1000 * 1000 + 7;
const int nax = 2e5 + 5;
int a[nax];
int c[nax];
int seg[nax << 2];
pair<int, int> b[nax];
pair<pair<int, int>, int> q[nax];
int ans[nax];
void update(int p, int l, int r, int i, int v) {
  if (i < l || i > r) return;
  seg[p] += v;
  if (l == r) {
    return;
  }
  int m = l + r >> 1;
  update(p << 1, l, m, i, v);
  update(p << 1 | 1, m + 1, r, i, v);
}
int get(int p, int l, int r, int pos) {
  int m = l + r >> 1;
  if (l == r) return l;
  if (seg[p << 1] < pos) return get(p << 1 | 1, m + 1, r, pos - seg[p << 1]);
  return get(p << 1, l, m, pos);
}
void solve() {
  int n, m;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> b[i].first;
    c[i] = b[i].first;
    b[i].first *= -1;
    b[i].second = i;
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    a[i + 1] = b[i].second;
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> q[i].first.first >> q[i].first.second;
    q[i].second = i;
  }
  sort(q, q + m);
  int done = 0;
  for (int i = 0; i < m; i++) {
    for (int j = done + 1; j <= q[i].first.first; j++) {
      update(1, 0, n - 1, a[j], 1);
    }
    done = max(done, q[i].first.first);
    int pos = get(1, 0, n - 1, q[i].first.second);
    ans[q[i].second] = c[pos];
  }
  for (int j = 0; j < m; j++) {
    cout << ans[j] << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout << fixed << setprecision(13);
  int t = 1;
  while (t--) {
    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;
template <class T>
void dbs(string str, T t) {
  cout << str << " : " << t << "\n";
}
template <class T, class... second>
void dbs(string str, T t, second... s) {
  long long idx = str.find(',');
  cout << str.substr(0, idx) << " : " << t << ", ";
  dbs(str.substr(idx + 1), s...);
}
template <class T>
void prc(T a, T b) {
  cout << "[";
  for (T i = a; i != b; ++i) {
    if (i != a) cout << ", ";
    cout << *i;
  }
  cout << "]\n";
}
template <class T>
void prall(T a) {
  prc(a.begin(), a.end());
}
const int inf = 1000 * 1000 * 1000 + 9;
const long long infl = 1ll * inf * inf;
const int mod = 1000 * 1000 * 1000 + 7;
const int nax = 2e5 + 5;
int a[nax];
int c[nax];
int seg[nax << 2];
pair<int, int> b[nax];
pair<pair<int, int>, int> q[nax];
int ans[nax];
void update(int p, int l, int r, int i, int v) {
  if (i < l || i > r) return;
  seg[p] += v;
  if (l == r) {
    return;
  }
  int m = l + r >> 1;
  update(p << 1, l, m, i, v);
  update(p << 1 | 1, m + 1, r, i, v);
}
int get(int p, int l, int r, int pos) {
  int m = l + r >> 1;
  if (l == r) return l;
  if (seg[p << 1] < pos) return get(p << 1 | 1, m + 1, r, pos - seg[p << 1]);
  return get(p << 1, l, m, pos);
}
void solve() {
  int n, m;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> b[i].first;
    c[i] = b[i].first;
    b[i].first *= -1;
    b[i].second = i;
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    a[i + 1] = b[i].second;
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> q[i].first.first >> q[i].first.second;
    q[i].second = i;
  }
  sort(q, q + m);
  int done = 0;
  for (int i = 0; i < m; i++) {
    for (int j = done + 1; j <= q[i].first.first; j++) {
      update(1, 0, n - 1, a[j], 1);
    }
    done = max(done, q[i].first.first);
    int pos = get(1, 0, n - 1, q[i].first.second);
    ans[q[i].second] = c[pos];
  }
  for (int j = 0; j < m; j++) {
    cout << ans[j] << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout << fixed << setprecision(13);
  int t = 1;
  while (t--) {
    solve();
  }
}

```