#include <bits/stdc++.h>
using namespace std;
bool comp(pair<long long int, long long int> p1,
          pair<long long int, long long int> p2) {
  if (p1.first != p2.first) return p1.first > p2.first;
  return p1.second < p2.second;
}
long long int auxi[200001], a[200001];
vector<pair<long long int, long long int>> adj[200001];
long long int ffs(long long int x) {
  for (long long int i = 0; i <= 31; i++) {
    if (x & (1LL << i)) return i + 1;
  }
  return 30;
}
void update(long long int idx, long long int n) {
  long long int x = idx;
  for (;;) {
    auxi[x] += 1;
    long long int y = x;
    long long int z = ffs(x);
    z--;
    z = (1LL << z);
    long long int w = y + z;
    x = w;
    if (x > 200000) break;
  }
}
long long int qry(long long int idx) {
  long long int s = 0;
  while (idx > 0) {
    s += auxi[idx];
    long long int x = ffs(idx);
    x--;
    x = 1LL << x;
    idx -= x;
  }
  return s;
}
long long int query(long long int val) {
  long long int start = 1, end = 200000, idx = 0, mid;
  while (start <= end) {
    mid = (start + end) / 2;
    long long int x = qry(mid);
    if (x >= val) {
      idx = mid;
      end = mid - 1;
    } else if (x < val)
      start = mid + 1;
  }
  return idx;
}
void solve() {
  long long int i, j, k, l, m, n, o;
  cin >> n;
  memset(auxi, 0, sizeof(auxi));
  vector<pair<long long int, long long int>> vp;
  for (i = 1; i <= n; i++) {
    cin >> j;
    a[i] = j;
    vp.push_back({j, i});
  }
  sort(vp.begin(), vp.end(), comp);
  cin >> m;
  vector<long long int> ans(m + 1);
  for (i = 0; i < m; i++) {
    cin >> j >> k;
    adj[j].push_back({k, i});
  }
  for (i = 1; i <= n; i++) {
    long long int z = vp[i - 1].second;
    update(z, n);
    for (auto it : adj[i]) {
      long long int y = query(it.first);
      ans[it.second] = a[y];
    }
  }
  for (i = 0; i < m; i++) cout << ans[i] << endl;
}
int main() {
  long long int t = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int i = 1;
  while (t--) {
    solve();
    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;
bool comp(pair<long long int, long long int> p1,
          pair<long long int, long long int> p2) {
  if (p1.first != p2.first) return p1.first > p2.first;
  return p1.second < p2.second;
}
long long int auxi[200001], a[200001];
vector<pair<long long int, long long int>> adj[200001];
long long int ffs(long long int x) {
  for (long long int i = 0; i <= 31; i++) {
    if (x & (1LL << i)) return i + 1;
  }
  return 30;
}
void update(long long int idx, long long int n) {
  long long int x = idx;
  for (;;) {
    auxi[x] += 1;
    long long int y = x;
    long long int z = ffs(x);
    z--;
    z = (1LL << z);
    long long int w = y + z;
    x = w;
    if (x > 200000) break;
  }
}
long long int qry(long long int idx) {
  long long int s = 0;
  while (idx > 0) {
    s += auxi[idx];
    long long int x = ffs(idx);
    x--;
    x = 1LL << x;
    idx -= x;
  }
  return s;
}
long long int query(long long int val) {
  long long int start = 1, end = 200000, idx = 0, mid;
  while (start <= end) {
    mid = (start + end) / 2;
    long long int x = qry(mid);
    if (x >= val) {
      idx = mid;
      end = mid - 1;
    } else if (x < val)
      start = mid + 1;
  }
  return idx;
}
void solve() {
  long long int i, j, k, l, m, n, o;
  cin >> n;
  memset(auxi, 0, sizeof(auxi));
  vector<pair<long long int, long long int>> vp;
  for (i = 1; i <= n; i++) {
    cin >> j;
    a[i] = j;
    vp.push_back({j, i});
  }
  sort(vp.begin(), vp.end(), comp);
  cin >> m;
  vector<long long int> ans(m + 1);
  for (i = 0; i < m; i++) {
    cin >> j >> k;
    adj[j].push_back({k, i});
  }
  for (i = 1; i <= n; i++) {
    long long int z = vp[i - 1].second;
    update(z, n);
    for (auto it : adj[i]) {
      long long int y = query(it.first);
      ans[it.second] = a[y];
    }
  }
  for (i = 0; i < m; i++) cout << ans[i] << endl;
}
int main() {
  long long int t = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int i = 1;
  while (t--) {
    solve();
    i++;
  }
  return 0;
}

```