#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)2e5 + 10;
int v[maxn];
int k[maxn], pos[maxn];
int num[maxn];
bool cmp(int a, int b) { return k[a] < k[b]; }
int t[4 * maxn];
void change(int it, int l, int r, int pos) {
  t[it]++;
  if (l != r) {
    int m = (l + r) >> 1;
    if (pos <= m) {
      change(2 * it, l, m, pos);
    } else {
      change(2 * it + 1, m + 1, r, pos);
    }
  }
}
int go(int it, int l, int r, int cnt) {
  if (l == r) {
    return l;
  } else {
    int m = (l + r) >> 1;
    if (t[2 * it] > cnt) {
      return go(2 * it, l, m, cnt);
    } else {
      return go(2 * it + 1, m + 1, r, cnt - t[2 * it]);
    }
  }
}
int res[maxn];
int main() {
  int n, m;
  cin >> n;
  vector<pair<int, int> > g;
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
    g.push_back(make_pair(v[i], -i));
  }
  sort(g.rbegin(), g.rend());
  cin >> m;
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &k[i], &pos[i]);
    pos[i]--;
    num[i] = i;
  }
  sort(num, num + m, cmp);
  int itr = 0;
  for (int ii = 0; ii < m; ii++) {
    int i = num[ii];
    while (itr < k[i]) {
      int p = -g[itr].second;
      change(1, 0, n - 1, p);
      itr++;
    }
    int ans = go(1, 0, n - 1, pos[i]);
    res[i] = v[ans];
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", res[i]);
  }
}

### Prompt

In cpp, your task is to solve 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 = (int)2e5 + 10;
int v[maxn];
int k[maxn], pos[maxn];
int num[maxn];
bool cmp(int a, int b) { return k[a] < k[b]; }
int t[4 * maxn];
void change(int it, int l, int r, int pos) {
  t[it]++;
  if (l != r) {
    int m = (l + r) >> 1;
    if (pos <= m) {
      change(2 * it, l, m, pos);
    } else {
      change(2 * it + 1, m + 1, r, pos);
    }
  }
}
int go(int it, int l, int r, int cnt) {
  if (l == r) {
    return l;
  } else {
    int m = (l + r) >> 1;
    if (t[2 * it] > cnt) {
      return go(2 * it, l, m, cnt);
    } else {
      return go(2 * it + 1, m + 1, r, cnt - t[2 * it]);
    }
  }
}
int res[maxn];
int main() {
  int n, m;
  cin >> n;
  vector<pair<int, int> > g;
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
    g.push_back(make_pair(v[i], -i));
  }
  sort(g.rbegin(), g.rend());
  cin >> m;
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &k[i], &pos[i]);
    pos[i]--;
    num[i] = i;
  }
  sort(num, num + m, cmp);
  int itr = 0;
  for (int ii = 0; ii < m; ii++) {
    int i = num[ii];
    while (itr < k[i]) {
      int p = -g[itr].second;
      change(1, 0, n - 1, p);
      itr++;
    }
    int ans = go(1, 0, n - 1, pos[i]);
    res[i] = v[ans];
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", res[i]);
  }
}

```