#include <bits/stdc++.h>
using namespace std;
const int Maxn = 200005;
const int Maxs = 1048576;
int n;
int a[Maxn];
pair<int, int> seq[Maxn];
int q;
int qa[Maxn], qb[Maxn];
pair<int, int> quer[Maxn];
int res[Maxn];
int st[Maxs];
void Add(int v, int l, int r, int x) {
  st[v]++;
  if (l < r) {
    int m = l + r >> 1;
    if (x <= m)
      Add(2 * v, l, m, x);
    else
      Add(2 * v + 1, m + 1, r, x);
  }
}
int Get(int v, int l, int r, int ind) {
  if (l == r)
    return a[l];
  else {
    int m = l + r >> 1;
    if (ind <= st[2 * v])
      return Get(2 * v, l, m, ind);
    else
      return Get(2 * v + 1, m + 1, r, ind - st[2 * v]);
  }
}
bool Less(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second < b.second;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    seq[i] = pair<int, int>(a[i], i);
  }
  sort(seq, seq + n, Less);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &qa[i], &qb[i]);
    quer[i] = pair<int, int>(qa[i], i);
  }
  sort(quer, quer + q);
  int pnt = 0;
  for (int i = 0; i < q; i++) {
    int ind = quer[i].second;
    while (pnt < qa[ind]) {
      int ind2 = seq[pnt].second;
      Add(1, 0, n - 1, ind2);
      pnt++;
    }
    res[ind] = Get(1, 0, n - 1, qb[ind]);
  }
  for (int i = 0; i < q; i++) printf("%d\n", res[i]);
  return 0;
}

### Prompt

Create a solution in cpp for 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 int Maxn = 200005;
const int Maxs = 1048576;
int n;
int a[Maxn];
pair<int, int> seq[Maxn];
int q;
int qa[Maxn], qb[Maxn];
pair<int, int> quer[Maxn];
int res[Maxn];
int st[Maxs];
void Add(int v, int l, int r, int x) {
  st[v]++;
  if (l < r) {
    int m = l + r >> 1;
    if (x <= m)
      Add(2 * v, l, m, x);
    else
      Add(2 * v + 1, m + 1, r, x);
  }
}
int Get(int v, int l, int r, int ind) {
  if (l == r)
    return a[l];
  else {
    int m = l + r >> 1;
    if (ind <= st[2 * v])
      return Get(2 * v, l, m, ind);
    else
      return Get(2 * v + 1, m + 1, r, ind - st[2 * v]);
  }
}
bool Less(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second < b.second;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    seq[i] = pair<int, int>(a[i], i);
  }
  sort(seq, seq + n, Less);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &qa[i], &qb[i]);
    quer[i] = pair<int, int>(qa[i], i);
  }
  sort(quer, quer + q);
  int pnt = 0;
  for (int i = 0; i < q; i++) {
    int ind = quer[i].second;
    while (pnt < qa[ind]) {
      int ind2 = seq[pnt].second;
      Add(1, 0, n - 1, ind2);
      pnt++;
    }
    res[ind] = Get(1, 0, n - 1, qb[ind]);
  }
  for (int i = 0; i < q; i++) printf("%d\n", res[i]);
  return 0;
}

```