#include <bits/stdc++.h>
using namespace std;
long long minimum(long long x, long long y) { return min(x, y); }
long long maximum(long long x, long long y) { return max(x, y); }
int main() {
  long long n;
  cin >> n;
  vector<long long> a(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<vector<vector<long long> > > dp(n + 1,
                                         vector<vector<long long> >(n + 1));
  vector<long long> z;
  for (int i = 1; i <= n; i++) {
    z.push_back(a[i]);
    dp[i][i] = z;
  }
  vector<long long> z1;
  z1.push_back(a[1]);
  for (int i = 2; i <= n; i++) {
    z1[0] = maximum(z1[0], a[i]);
    dp[1][i] = z1;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      vector<long long> a1 = dp[i - 1][j - 1];
      vector<long long> a2 = dp[i][j - 1];
      a1.push_back(a[j]);
      long long sum1 = 0;
      long long sum2 = 0;
      for (int k = 0; k < i; k++) {
        sum1 += a1[k];
        sum2 += a2[k];
      }
      if (sum1 > sum2)
        dp[i][j] = a1;
      else if (sum2 > sum1)
        dp[i][j] = a2;
      else {
        int k = 0;
        bool f1 = false;
        bool f2 = false;
        while (k < i) {
          if (a1[k] == a2[k])
            k++;
          else if (a1[k] < a2[k]) {
            f1 = true;
            break;
          } else {
            f2 = true;
            break;
          }
        }
        if (f1)
          dp[i][j] = a1;
        else
          dp[i][j] = a2;
      }
    }
  }
  long long m;
  cin >> m;
  while (m--) {
    long long k, pos;
    cin >> k >> pos;
    cout << dp[k][n][pos - 1] << endl;
  }
}

### 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;
long long minimum(long long x, long long y) { return min(x, y); }
long long maximum(long long x, long long y) { return max(x, y); }
int main() {
  long long n;
  cin >> n;
  vector<long long> a(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<vector<vector<long long> > > dp(n + 1,
                                         vector<vector<long long> >(n + 1));
  vector<long long> z;
  for (int i = 1; i <= n; i++) {
    z.push_back(a[i]);
    dp[i][i] = z;
  }
  vector<long long> z1;
  z1.push_back(a[1]);
  for (int i = 2; i <= n; i++) {
    z1[0] = maximum(z1[0], a[i]);
    dp[1][i] = z1;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      vector<long long> a1 = dp[i - 1][j - 1];
      vector<long long> a2 = dp[i][j - 1];
      a1.push_back(a[j]);
      long long sum1 = 0;
      long long sum2 = 0;
      for (int k = 0; k < i; k++) {
        sum1 += a1[k];
        sum2 += a2[k];
      }
      if (sum1 > sum2)
        dp[i][j] = a1;
      else if (sum2 > sum1)
        dp[i][j] = a2;
      else {
        int k = 0;
        bool f1 = false;
        bool f2 = false;
        while (k < i) {
          if (a1[k] == a2[k])
            k++;
          else if (a1[k] < a2[k]) {
            f1 = true;
            break;
          } else {
            f2 = true;
            break;
          }
        }
        if (f1)
          dp[i][j] = a1;
        else
          dp[i][j] = a2;
      }
    }
  }
  long long m;
  cin >> m;
  while (m--) {
    long long k, pos;
    cin >> k >> pos;
    cout << dp[k][n][pos - 1] << endl;
  }
}

```