#include <bits/stdc++.h>
using namespace std;
inline long long gcd(long long a, long long b) {
  return b > 0 ? gcd(b, a % b) : a;
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
inline long long qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a;
    }
    a *= a;
    n >>= 1;
  }
  return ans;
}
inline long long _qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a % 1000000007;
      ans %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    n >>= 1;
  }
  return ans % 1000000007;
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      f = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const long long maxnum = 105;
long long a[maxnum];
bool vis[maxnum];
vector<long long> b[maxnum];
signed main() {
  long long n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  priority_queue<long long, vector<long long>, greater<long long>> que;
  long long sum = 0;
  for (long long i = 1; i <= n; i++) sum += a[i], que.push(a[i]);
  b[n].push_back(0);
  for (long long i = 1; i <= n; i++) b[n].push_back(a[i]);
  for (long long k = n - 1; k >= 1; k--) {
    long long tar = que.top();
    que.pop();
    for (long long i = n; i >= 1; i--) {
      if (vis[i]) continue;
      if (tar == a[i]) {
        vis[i] = true;
        break;
      }
    }
    b[k].push_back(0);
    for (long long i = 1; i <= n; i++)
      if (!vis[i]) b[k].push_back(a[i]);
  }
  long long q = read();
  while (q--) {
    long long k = read(), idx = read();
    cout << b[k][idx] << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
inline long long gcd(long long a, long long b) {
  return b > 0 ? gcd(b, a % b) : a;
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
inline long long qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a;
    }
    a *= a;
    n >>= 1;
  }
  return ans;
}
inline long long _qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a % 1000000007;
      ans %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    n >>= 1;
  }
  return ans % 1000000007;
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      f = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const long long maxnum = 105;
long long a[maxnum];
bool vis[maxnum];
vector<long long> b[maxnum];
signed main() {
  long long n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  priority_queue<long long, vector<long long>, greater<long long>> que;
  long long sum = 0;
  for (long long i = 1; i <= n; i++) sum += a[i], que.push(a[i]);
  b[n].push_back(0);
  for (long long i = 1; i <= n; i++) b[n].push_back(a[i]);
  for (long long k = n - 1; k >= 1; k--) {
    long long tar = que.top();
    que.pop();
    for (long long i = n; i >= 1; i--) {
      if (vis[i]) continue;
      if (tar == a[i]) {
        vis[i] = true;
        break;
      }
    }
    b[k].push_back(0);
    for (long long i = 1; i <= n; i++)
      if (!vis[i]) b[k].push_back(a[i]);
  }
  long long q = read();
  while (q--) {
    long long k = read(), idx = read();
    cout << b[k][idx] << endl;
  }
  return 0;
}

```