#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long rand(long long x, long long y) {
  return (rng() % (y + 1 - x)) + x;
}
string inline to_string(char c) {
  string second(1, c);
  return second;
}
template <typename T>
inline T gcd(T a, T b) {
  return a == 0 ? b : gcd(b % a, a);
}
long long int n, m;
vector<pair<long long int, long long int>> A;
vector<long long int> o;
vector<vector<long long int>> co;
map<long long int, long long int> mp;
vector<long long int> conf;
struct fen {
  int fw[(200006)];
  fen() { memset((fw), 0, sizeof((fw))); }
  void update(long long int x, long long int nval) {
    for (; x < (200006); x += x & (-x)) fw[x] += nval;
  }
  long long int sum(long long int x) {
    long long int res = 0;
    for (; x; x -= x & (-x)) res += fw[x];
    return res;
  }
} fen;
long long int ret(long long int x) {
  long long int st = 0, en = (200006), mid = 0;
  while (en - st > 1) {
    mid = (st + en) >> 1;
    if (fen.sum(mid) >= x)
      en = mid;
    else
      st = mid;
  }
  assert(en != (200006));
  return o[en - 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  A.resize(n);
  o.resize(n);
  for (long long int i = (0); i <= (long long int)(n - 1); ++i)
    cin >> A[i].first, A[i].second = i, o[i] = A[i].first;
  sort((A).begin(), (A).end());
  for (long long int i = (0); i <= (long long int)(n - 1); ++i) {
    if (i == 0 || A[i].first != A[i - 1].first) {
      co.push_back(vector<long long int>());
    }
    co.back().push_back(A[i].second);
    mp[A[i].first] = co.size() - 1;
  }
  cin >> m;
  vector<pair<long long int, pair<long long int, long long int>>> Q;
  vector<long long int> ans(m, 0);
  for (long long int i = (0); i <= (long long int)(m - 1); ++i) {
    long long int k, pos;
    cin >> k >> pos;
    Q.emplace_back(i, pair<long long int, long long int>(k, pos));
  }
  sort((Q).begin(), (Q).end(),
       [](pair<long long int, pair<long long int, long long int>> x,
          pair<long long int, pair<long long int, long long int>> y) {
         return x.second.first < y.second.first;
       });
  sort((A).begin(), (A).end(),
       [](pair<long long int, long long int> x,
          pair<long long int, long long int> y) {
         if (x.first ^ y.first)
           return x.first < y.first;
         else
           return x.second > y.second;
       });
  for (long long int ii = (0); ii <= (long long int)(m - 1); ++ii) {
    long long int k = Q[ii].second.first, pos = Q[ii].second.second;
    while (n - A.size() < k) {
      fen.update(A.back().second + 1, 1);
      A.pop_back();
    }
    assert(fen.sum((200006) - 1) == k);
    ans[Q[ii].first] = ret(pos);
  }
  for (auto i : ans) cout << i << '\n';
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long rand(long long x, long long y) {
  return (rng() % (y + 1 - x)) + x;
}
string inline to_string(char c) {
  string second(1, c);
  return second;
}
template <typename T>
inline T gcd(T a, T b) {
  return a == 0 ? b : gcd(b % a, a);
}
long long int n, m;
vector<pair<long long int, long long int>> A;
vector<long long int> o;
vector<vector<long long int>> co;
map<long long int, long long int> mp;
vector<long long int> conf;
struct fen {
  int fw[(200006)];
  fen() { memset((fw), 0, sizeof((fw))); }
  void update(long long int x, long long int nval) {
    for (; x < (200006); x += x & (-x)) fw[x] += nval;
  }
  long long int sum(long long int x) {
    long long int res = 0;
    for (; x; x -= x & (-x)) res += fw[x];
    return res;
  }
} fen;
long long int ret(long long int x) {
  long long int st = 0, en = (200006), mid = 0;
  while (en - st > 1) {
    mid = (st + en) >> 1;
    if (fen.sum(mid) >= x)
      en = mid;
    else
      st = mid;
  }
  assert(en != (200006));
  return o[en - 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  A.resize(n);
  o.resize(n);
  for (long long int i = (0); i <= (long long int)(n - 1); ++i)
    cin >> A[i].first, A[i].second = i, o[i] = A[i].first;
  sort((A).begin(), (A).end());
  for (long long int i = (0); i <= (long long int)(n - 1); ++i) {
    if (i == 0 || A[i].first != A[i - 1].first) {
      co.push_back(vector<long long int>());
    }
    co.back().push_back(A[i].second);
    mp[A[i].first] = co.size() - 1;
  }
  cin >> m;
  vector<pair<long long int, pair<long long int, long long int>>> Q;
  vector<long long int> ans(m, 0);
  for (long long int i = (0); i <= (long long int)(m - 1); ++i) {
    long long int k, pos;
    cin >> k >> pos;
    Q.emplace_back(i, pair<long long int, long long int>(k, pos));
  }
  sort((Q).begin(), (Q).end(),
       [](pair<long long int, pair<long long int, long long int>> x,
          pair<long long int, pair<long long int, long long int>> y) {
         return x.second.first < y.second.first;
       });
  sort((A).begin(), (A).end(),
       [](pair<long long int, long long int> x,
          pair<long long int, long long int> y) {
         if (x.first ^ y.first)
           return x.first < y.first;
         else
           return x.second > y.second;
       });
  for (long long int ii = (0); ii <= (long long int)(m - 1); ++ii) {
    long long int k = Q[ii].second.first, pos = Q[ii].second.second;
    while (n - A.size() < k) {
      fen.update(A.back().second + 1, 1);
      A.pop_back();
    }
    assert(fen.sum((200006) - 1) == k);
    ans[Q[ii].first] = ret(pos);
  }
  for (auto i : ans) cout << i << '\n';
}

```