#include <bits/stdc++.h>
using namespace std;
const int MAX = 200010;
struct wavelet_tree {
  int lo, hi;
  wavelet_tree *l, *r;
  vector<int> b;
  wavelet_tree(int *from, int *to, int x, int y) {
    lo = x, hi = y;
    if (lo == hi or from >= to) return;
    int mid = (lo + hi) / 2;
    auto f = [mid](int x) { return x <= mid; };
    b.reserve(to - from + 1);
    b.push_back(0);
    for (auto it = from; it != to; it++) b.push_back(b.back() + f(*it));
    auto pivot = stable_partition(from, to, f);
    l = new wavelet_tree(from, pivot, lo, mid);
    r = new wavelet_tree(pivot, to, mid + 1, hi);
  }
  int kth(int l, int r, int k) {
    if (l > r) return 0;
    if (lo == hi) return lo;
    int inLeft = b[r] - b[l - 1];
    int lb = b[l - 1];
    int rb = b[r];
    if (k <= inLeft) return this->l->kth(lb + 1, rb, k);
    return this->r->kth(l - lb, r - rb, k - inLeft);
  }
  ~wavelet_tree() {
    delete l;
    delete r;
  }
};
vector<pair<int, int> > V;
int ara[MAX];
int bra[MAX];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ara[i];
    V.push_back({-ara[i], i});
  }
  sort(V.begin(), V.end());
  for (int i = 1; i <= n; i++) bra[i] = V[i - 1].second;
  wavelet_tree *Tree = new wavelet_tree(bra + 1, bra + n + 1, 1, 2e5);
  cin >> m;
  int p, a, b;
  for (int i = 1; i <= m; i++) {
    cin >> a >> b;
    p = Tree->kth(1, a, b);
    cout << ara[p] << '\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>
using namespace std;
const int MAX = 200010;
struct wavelet_tree {
  int lo, hi;
  wavelet_tree *l, *r;
  vector<int> b;
  wavelet_tree(int *from, int *to, int x, int y) {
    lo = x, hi = y;
    if (lo == hi or from >= to) return;
    int mid = (lo + hi) / 2;
    auto f = [mid](int x) { return x <= mid; };
    b.reserve(to - from + 1);
    b.push_back(0);
    for (auto it = from; it != to; it++) b.push_back(b.back() + f(*it));
    auto pivot = stable_partition(from, to, f);
    l = new wavelet_tree(from, pivot, lo, mid);
    r = new wavelet_tree(pivot, to, mid + 1, hi);
  }
  int kth(int l, int r, int k) {
    if (l > r) return 0;
    if (lo == hi) return lo;
    int inLeft = b[r] - b[l - 1];
    int lb = b[l - 1];
    int rb = b[r];
    if (k <= inLeft) return this->l->kth(lb + 1, rb, k);
    return this->r->kth(l - lb, r - rb, k - inLeft);
  }
  ~wavelet_tree() {
    delete l;
    delete r;
  }
};
vector<pair<int, int> > V;
int ara[MAX];
int bra[MAX];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ara[i];
    V.push_back({-ara[i], i});
  }
  sort(V.begin(), V.end());
  for (int i = 1; i <= n; i++) bra[i] = V[i - 1].second;
  wavelet_tree *Tree = new wavelet_tree(bra + 1, bra + n + 1, 1, 2e5);
  cin >> m;
  int p, a, b;
  for (int i = 1; i <= m; i++) {
    cin >> a >> b;
    p = Tree->kth(1, a, b);
    cout << ara[p] << '\n';
  }
  return 0;
}

```