#include <bits/stdc++.h>
using namespace std;
const int maxn = (1 << 18);
int t1[maxn];
int t2[maxn];
void add(int *t, int p, int val) {
  for (; p < maxn; p += (p & -p)) t[p] += val;
}
int fnd(int k, int li) {
  int now = 0;
  for (int i = 17; i >= 0; --i) {
    if (t1[now + (1 << i)] + min(li, t2[now + (1 << i)]) < k) {
      k -= (t1[now + (1 << i)] + min(li, t2[now + (1 << i)]));
      li -= min(li, t2[now + (1 << i)]);
      now += (1 << i);
    }
  }
  return now + 1;
}
vector<int> pos[maxn];
int a[maxn], b[maxn];
int ans[maxn];
struct Node {
  int k, p;
  int no;
} node[maxn];
void addall(int *t, vector<int> &p, int val) {
  for (int i = 0; i < p.size(); ++i) add(t, p[i], val);
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  int m;
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> node[i].k >> node[i].p;
    node[i].no = i;
  }
  sort(node + 1, node + 1 + m,
       [](const Node &a, const Node &b) { return a.k < b.k; });
  for (int i = 1; i <= n; ++i) b[i] = a[i];
  sort(b + 1, b + 1 + n);
  int d = unique(b + 1, b + 1 + n) - (b + 1);
  for (int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + 1 + d, a[i]) - b;
  for (int i = 1; i <= n; ++i) pos[a[i]].push_back(i);
  int nowsz = 0;
  addall(t2, pos[d], 1);
  for (int q = 1; q <= m; ++q) {
    while (nowsz + pos[d].size() < node[q].k) {
      addall(t2, pos[d], -1);
      addall(t1, pos[d], 1);
      nowsz += pos[d].size();
      --d;
      addall(t2, pos[d], 1);
    }
    ans[node[q].no] = fnd(node[q].p, node[q].k - nowsz);
  }
  for (int q = 1; q <= m; ++q) {
    cout << b[a[ans[q]]] << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int maxn = (1 << 18);
int t1[maxn];
int t2[maxn];
void add(int *t, int p, int val) {
  for (; p < maxn; p += (p & -p)) t[p] += val;
}
int fnd(int k, int li) {
  int now = 0;
  for (int i = 17; i >= 0; --i) {
    if (t1[now + (1 << i)] + min(li, t2[now + (1 << i)]) < k) {
      k -= (t1[now + (1 << i)] + min(li, t2[now + (1 << i)]));
      li -= min(li, t2[now + (1 << i)]);
      now += (1 << i);
    }
  }
  return now + 1;
}
vector<int> pos[maxn];
int a[maxn], b[maxn];
int ans[maxn];
struct Node {
  int k, p;
  int no;
} node[maxn];
void addall(int *t, vector<int> &p, int val) {
  for (int i = 0; i < p.size(); ++i) add(t, p[i], val);
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  int m;
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> node[i].k >> node[i].p;
    node[i].no = i;
  }
  sort(node + 1, node + 1 + m,
       [](const Node &a, const Node &b) { return a.k < b.k; });
  for (int i = 1; i <= n; ++i) b[i] = a[i];
  sort(b + 1, b + 1 + n);
  int d = unique(b + 1, b + 1 + n) - (b + 1);
  for (int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + 1 + d, a[i]) - b;
  for (int i = 1; i <= n; ++i) pos[a[i]].push_back(i);
  int nowsz = 0;
  addall(t2, pos[d], 1);
  for (int q = 1; q <= m; ++q) {
    while (nowsz + pos[d].size() < node[q].k) {
      addall(t2, pos[d], -1);
      addall(t1, pos[d], 1);
      nowsz += pos[d].size();
      --d;
      addall(t2, pos[d], 1);
    }
    ans[node[q].no] = fnd(node[q].p, node[q].k - nowsz);
  }
  for (int q = 1; q <= m; ++q) {
    cout << b[a[ans[q]]] << endl;
  }
  return 0;
}

```