#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 9;
const long long mod = 1e9 + 7;
vector<bool> prime(MAX, 1);
vector<int> spf(MAX, 1);
vector<int> primes;
void sieve() {
  prime[0] = prime[1] = 0;
  spf[2] = 2;
  for (long long i = 4; i < MAX; i += 2) {
    spf[i] = 2;
    prime[i] = 0;
  }
  primes.push_back(2);
  for (long long i = 3; i < MAX; i += 2) {
    if (prime[i]) {
      primes.push_back(i);
      spf[i] = i;
      for (long long j = i * i; j < MAX; j += i) {
        prime[j] = 0;
        if (spf[j] == 1) {
          spf[j] = i;
        }
      }
    }
  }
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a;
    }
    a = a * a;
    b = b >> 1;
  }
  return res;
}
long long power(long long a, long long b, long long m) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % m;
    }
    a = (a * a) % m;
    b = b >> 1;
  }
  return res % m;
}
void virtual_main() {}
void real_main() {
  int n;
  cin >> n;
  vector<int> v(n), a(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    a[i] = v[i];
  }
  sort(v.begin(), v.end());
  int m;
  cin >> m;
  while (m--) {
    int k, pos;
    cin >> k >> pos;
    vector<int> values;
    for (int i = n - 1; i >= n - k; i--) {
      values.push_back(v[i]);
    }
    reverse(values.begin(), values.end());
    vector<int> ans;
    vector<bool> mark(n, 0);
    int last = 0;
    for (int i = 0; i < (int)values.size(); i++) {
      int x = values[i];
      bool say = 0;
      for (int j = last; j < n; j++) {
        if (a[j] == x && !mark[j]) {
          say = 1;
          ans.push_back(j);
          mark[j] = 1;
          last = j;
          break;
        }
      }
      if (!say) {
        for (int j = 0; j <= last; j++) {
          if (a[j] == x && !mark[j]) {
            say = 1;
            mark[j] = 1;
            last = j;
            ans.push_back(j);
            break;
          }
        }
      }
    }
    sort(ans.begin(), ans.end());
    cout << a[ans[pos - 1]] << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  virtual_main();
  int test_cases = 1;
  for (int i = 1; i <= test_cases; i++) {
    real_main();
  }
  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;
const int MAX = 1e5 + 9;
const long long mod = 1e9 + 7;
vector<bool> prime(MAX, 1);
vector<int> spf(MAX, 1);
vector<int> primes;
void sieve() {
  prime[0] = prime[1] = 0;
  spf[2] = 2;
  for (long long i = 4; i < MAX; i += 2) {
    spf[i] = 2;
    prime[i] = 0;
  }
  primes.push_back(2);
  for (long long i = 3; i < MAX; i += 2) {
    if (prime[i]) {
      primes.push_back(i);
      spf[i] = i;
      for (long long j = i * i; j < MAX; j += i) {
        prime[j] = 0;
        if (spf[j] == 1) {
          spf[j] = i;
        }
      }
    }
  }
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a;
    }
    a = a * a;
    b = b >> 1;
  }
  return res;
}
long long power(long long a, long long b, long long m) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % m;
    }
    a = (a * a) % m;
    b = b >> 1;
  }
  return res % m;
}
void virtual_main() {}
void real_main() {
  int n;
  cin >> n;
  vector<int> v(n), a(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    a[i] = v[i];
  }
  sort(v.begin(), v.end());
  int m;
  cin >> m;
  while (m--) {
    int k, pos;
    cin >> k >> pos;
    vector<int> values;
    for (int i = n - 1; i >= n - k; i--) {
      values.push_back(v[i]);
    }
    reverse(values.begin(), values.end());
    vector<int> ans;
    vector<bool> mark(n, 0);
    int last = 0;
    for (int i = 0; i < (int)values.size(); i++) {
      int x = values[i];
      bool say = 0;
      for (int j = last; j < n; j++) {
        if (a[j] == x && !mark[j]) {
          say = 1;
          ans.push_back(j);
          mark[j] = 1;
          last = j;
          break;
        }
      }
      if (!say) {
        for (int j = 0; j <= last; j++) {
          if (a[j] == x && !mark[j]) {
            say = 1;
            mark[j] = 1;
            last = j;
            ans.push_back(j);
            break;
          }
        }
      }
    }
    sort(ans.begin(), ans.end());
    cout << a[ans[pos - 1]] << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  virtual_main();
  int test_cases = 1;
  for (int i = 1; i <= test_cases; i++) {
    real_main();
  }
  return 0;
}

```