#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << '\n';
  err(++it, args...);
}
const long long inf = 1e18L + 5;
const long long mod = 1e9 + 7;
const long long SIZEN = 200005;
long long arr[SIZEN], tree[SIZEN * 4];
vector<pair<long long, long long> > sortedVec, vec2, ques;
vector<vector<long long> > answer;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first > b.first) return true;
  if (b.first > a.first) return false;
  if (a.second > b.second) return false;
  return true;
}
bool comp2(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first > b.first) return true;
  if (b.first > a.first) return false;
  return a.second < b.second;
}
void update(int v, int tl, int tr, int pos, int val) {
  if (tl == tr) {
    tree[v] += val;
  } else {
    int tm = (tl + tr) / 2;
    if (pos <= tm) {
      update(v * 2 + 1, tl, tm, pos, val);
    } else {
      update(v * 2 + 2, tm + 1, tr, pos, val);
    }
    tree[v] = tree[v * 2 + 1] + tree[v * 2 + 2];
  }
}
long long get(int v, int tl, int tr, int k) {
  if (tl == tr) {
    return tl;
  } else {
    int tm = (tl + tr) / 2;
    if (tree[v * 2 + 1] >= k) {
      return get(v * 2 + 1, tl, tm, k);
    } else {
      return get(v * 2 + 2, tm + 1, tr, k - tree[v * 2 + 1]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    sortedVec.push_back(make_pair(arr[i], i));
  }
  cin >> m;
  sort(sortedVec.begin(), sortedVec.end(), comp2);
  answer.resize(m, vector<long long>(3));
  for (int i = 0; i < m; i++) {
    int k, pos;
    cin >> k >> pos;
    ques.push_back(make_pair(k, pos));
    answer[i][0] = k;
    answer[i][1] = pos;
    answer[i][2] = i;
  }
  sort(answer.begin(), answer.end());
  vector<int> ans(m);
  int prev = 0;
  for (int i = 0; i < answer.size(); i++) {
    for (int j = prev; j < answer[i][0]; j++) {
      update(0, 0, n, sortedVec[j].second, 1);
      prev++;
    }
    ans[answer[i][2]] = arr[get(0, 0, n, answer[i][1])];
  }
  for (int i = 0; i < answer.size(); i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}

### Prompt

Generate a CPP solution to 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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << '\n';
  err(++it, args...);
}
const long long inf = 1e18L + 5;
const long long mod = 1e9 + 7;
const long long SIZEN = 200005;
long long arr[SIZEN], tree[SIZEN * 4];
vector<pair<long long, long long> > sortedVec, vec2, ques;
vector<vector<long long> > answer;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first > b.first) return true;
  if (b.first > a.first) return false;
  if (a.second > b.second) return false;
  return true;
}
bool comp2(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first > b.first) return true;
  if (b.first > a.first) return false;
  return a.second < b.second;
}
void update(int v, int tl, int tr, int pos, int val) {
  if (tl == tr) {
    tree[v] += val;
  } else {
    int tm = (tl + tr) / 2;
    if (pos <= tm) {
      update(v * 2 + 1, tl, tm, pos, val);
    } else {
      update(v * 2 + 2, tm + 1, tr, pos, val);
    }
    tree[v] = tree[v * 2 + 1] + tree[v * 2 + 2];
  }
}
long long get(int v, int tl, int tr, int k) {
  if (tl == tr) {
    return tl;
  } else {
    int tm = (tl + tr) / 2;
    if (tree[v * 2 + 1] >= k) {
      return get(v * 2 + 1, tl, tm, k);
    } else {
      return get(v * 2 + 2, tm + 1, tr, k - tree[v * 2 + 1]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    sortedVec.push_back(make_pair(arr[i], i));
  }
  cin >> m;
  sort(sortedVec.begin(), sortedVec.end(), comp2);
  answer.resize(m, vector<long long>(3));
  for (int i = 0; i < m; i++) {
    int k, pos;
    cin >> k >> pos;
    ques.push_back(make_pair(k, pos));
    answer[i][0] = k;
    answer[i][1] = pos;
    answer[i][2] = i;
  }
  sort(answer.begin(), answer.end());
  vector<int> ans(m);
  int prev = 0;
  for (int i = 0; i < answer.size(); i++) {
    for (int j = prev; j < answer[i][0]; j++) {
      update(0, 0, n, sortedVec[j].second, 1);
      prev++;
    }
    ans[answer[i][2]] = arr[get(0, 0, n, answer[i][1])];
  }
  for (int i = 0; i < answer.size(); i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}

```