#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first != b.first) {
    return a.first > b.first;
  } else {
    return a.second < b.second;
  }
}
const int N = 2e5 + 10;
int t[N], n, on[N], ans[N];
void update(int v) {
  for (; v <= n; v += v & (-v)) {
    t[v]++;
  }
}
int get(int v) {
  int res = 0;
  for (; v > 0; v -= v & (-v)) {
    res += t[v];
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<pair<int, int>> arr;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    on[i + 1] = x;
    arr.push_back({x, i + 1});
  }
  sort((arr).begin(), (arr).end(), cmp);
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int>> que;
  for (int i = 0; i < m; i++) {
    int k, pos;
    cin >> k >> pos;
    que.push_back(make_pair(make_pair(k, pos), i));
  }
  sort((que).begin(), (que).end());
  int ptr = 0;
  for (int i = 0; i < n; i++) {
    update(arr[i].second);
    while (ptr < m && que[ptr].first.first == i + 1) {
      int need = que[ptr].first.second;
      int l = 0, r = n + 1;
      while (r - l > 1) {
        int mid = (l + r) / 2;
        if (get(mid) >= need) {
          r = mid;
        } else {
          l = mid;
        }
      }
      ans[que[ptr].second] = on[r];
      ptr++;
    }
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << '\n';
  }
}

### 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;
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first != b.first) {
    return a.first > b.first;
  } else {
    return a.second < b.second;
  }
}
const int N = 2e5 + 10;
int t[N], n, on[N], ans[N];
void update(int v) {
  for (; v <= n; v += v & (-v)) {
    t[v]++;
  }
}
int get(int v) {
  int res = 0;
  for (; v > 0; v -= v & (-v)) {
    res += t[v];
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<pair<int, int>> arr;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    on[i + 1] = x;
    arr.push_back({x, i + 1});
  }
  sort((arr).begin(), (arr).end(), cmp);
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int>> que;
  for (int i = 0; i < m; i++) {
    int k, pos;
    cin >> k >> pos;
    que.push_back(make_pair(make_pair(k, pos), i));
  }
  sort((que).begin(), (que).end());
  int ptr = 0;
  for (int i = 0; i < n; i++) {
    update(arr[i].second);
    while (ptr < m && que[ptr].first.first == i + 1) {
      int need = que[ptr].first.second;
      int l = 0, r = n + 1;
      while (r - l > 1) {
        int mid = (l + r) / 2;
        if (get(mid) >= need) {
          r = mid;
        } else {
          l = mid;
        }
      }
      ans[que[ptr].second] = on[r];
      ptr++;
    }
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << '\n';
  }
}

```