#include <bits/stdc++.h>
using namespace std;
template <class T>
using vc = vector<T>;
template <class T>
using vvc = vc<vc<T>>;
template <class T>
void mkuni(vector<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
template <class T>
void print(T x, int suc = 1) {
  cout << x;
  if (suc == 1)
    cout << '\n';
  else
    cout << ' ';
}
template <class T>
void print(const vector<T>& v, int suc = 1) {
  for (int i = 0; i < v.size(); i++)
    print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
}
const int INF = 0x3f3f3f3f;
struct node {
  int v, pos;
  bool operator<(const node& a) const {
    if (v == a.v) return pos < a.pos;
    return v > a.v;
  }
};
const int maxn = 2e5 + 7;
vector<pair<int, int>> query[maxn];
int ans[maxn];
int n;
int cnt[maxn << 2];
void update(int p, int L = 1, int R = n, int o = 1) {
  cnt[o]++;
  if (L == R) return;
  int mid = L + R >> 1;
  if (p <= mid)
    update(p, L, mid, o << 1);
  else
    update(p, mid + 1, R, o << 1 | 1);
}
int getans(int k, int L = 1, int R = n, int o = 1) {
  if (L == R) return L;
  int mid = L + R >> 1;
  if (cnt[o << 1] >= k) return getans(k, L, mid, o << 1);
  return getans(k - cnt[o << 1], mid + 1, R, o << 1 | 1);
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  vector<int> a(n);
  vector<node> all;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    all.push_back(node{a[i], i});
  }
  sort(all.begin(), all.end());
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    pair<int, int> tmp;
    cin >> tmp.first >> tmp.second;
    query[tmp.first].push_back(make_pair(tmp.second, i));
  }
  for (int i = 1; i <= n; i++) {
    update(all[i - 1].pos + 1);
    ;
    sort(query[i].begin(), query[i].end());
    for (auto q : query[i]) ans[q.second] = a[getans(q.first) - 1];
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}

### 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;
template <class T>
using vc = vector<T>;
template <class T>
using vvc = vc<vc<T>>;
template <class T>
void mkuni(vector<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
template <class T>
void print(T x, int suc = 1) {
  cout << x;
  if (suc == 1)
    cout << '\n';
  else
    cout << ' ';
}
template <class T>
void print(const vector<T>& v, int suc = 1) {
  for (int i = 0; i < v.size(); i++)
    print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
}
const int INF = 0x3f3f3f3f;
struct node {
  int v, pos;
  bool operator<(const node& a) const {
    if (v == a.v) return pos < a.pos;
    return v > a.v;
  }
};
const int maxn = 2e5 + 7;
vector<pair<int, int>> query[maxn];
int ans[maxn];
int n;
int cnt[maxn << 2];
void update(int p, int L = 1, int R = n, int o = 1) {
  cnt[o]++;
  if (L == R) return;
  int mid = L + R >> 1;
  if (p <= mid)
    update(p, L, mid, o << 1);
  else
    update(p, mid + 1, R, o << 1 | 1);
}
int getans(int k, int L = 1, int R = n, int o = 1) {
  if (L == R) return L;
  int mid = L + R >> 1;
  if (cnt[o << 1] >= k) return getans(k, L, mid, o << 1);
  return getans(k - cnt[o << 1], mid + 1, R, o << 1 | 1);
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  vector<int> a(n);
  vector<node> all;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    all.push_back(node{a[i], i});
  }
  sort(all.begin(), all.end());
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    pair<int, int> tmp;
    cin >> tmp.first >> tmp.second;
    query[tmp.first].push_back(make_pair(tmp.second, i));
  }
  for (int i = 1; i <= n; i++) {
    update(all[i - 1].pos + 1);
    ;
    sort(query[i].begin(), query[i].end());
    for (auto q : query[i]) ans[q.second] = a[getans(q.first) - 1];
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}

```