#include <bits/stdc++.h>
using namespace std;
template <class T>
void byebye(T _rpl) {
  cout << _rpl << endl;
  exit(0);
}
int nextint() {
  int x;
  scanf("%d", &x);
  return x;
}
long long nextll() {
  long long x;
  scanf("%lld", &x);
  return x;
}
template <class T>
T emax(T& t1, T t2) {
  if (t1 < t2) t1 = t2;
  return t1;
}
template <class T>
T emin(T& t1, T t2) {
  if (t1 > t2) t1 = t2;
  return t1;
}
int read() {
  int f = 1, ret = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ret = ret * 10 + (c - '0');
    c = getchar();
  }
  return ret * f;
}
int n;
int a[105];
bool isin[105];
pair<int, int> s[105];
bool cmp(const pair<int, int>& a1, const pair<int, int>& a2) {
  if (a1.first != a2.first) return a1.first > a2.first;
  return a1.second < a2.second;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i)
    cin >> s[i].first, a[i] = s[i].first, s[i].second = i;
  sort(s, s + n, cmp);
  int q;
  cin >> q;
  while (q--) {
    int m, pos;
    cin >> m >> pos;
    memset(isin, 0, sizeof(isin));
    for (int i = 0; i < m; ++i) isin[s[i].second] = 1;
    int cnt = 0;
    for (int i = 0; i < n; ++i) {
      if (isin[i]) cnt++;
      if (cnt == pos) {
        cout << a[i] << endl;
        break;
      }
    }
  }
  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;
template <class T>
void byebye(T _rpl) {
  cout << _rpl << endl;
  exit(0);
}
int nextint() {
  int x;
  scanf("%d", &x);
  return x;
}
long long nextll() {
  long long x;
  scanf("%lld", &x);
  return x;
}
template <class T>
T emax(T& t1, T t2) {
  if (t1 < t2) t1 = t2;
  return t1;
}
template <class T>
T emin(T& t1, T t2) {
  if (t1 > t2) t1 = t2;
  return t1;
}
int read() {
  int f = 1, ret = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ret = ret * 10 + (c - '0');
    c = getchar();
  }
  return ret * f;
}
int n;
int a[105];
bool isin[105];
pair<int, int> s[105];
bool cmp(const pair<int, int>& a1, const pair<int, int>& a2) {
  if (a1.first != a2.first) return a1.first > a2.first;
  return a1.second < a2.second;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i)
    cin >> s[i].first, a[i] = s[i].first, s[i].second = i;
  sort(s, s + n, cmp);
  int q;
  cin >> q;
  while (q--) {
    int m, pos;
    cin >> m >> pos;
    memset(isin, 0, sizeof(isin));
    for (int i = 0; i < m; ++i) isin[s[i].second] = 1;
    int cnt = 0;
    for (int i = 0; i < n; ++i) {
      if (isin[i]) cnt++;
      if (cnt == pos) {
        cout << a[i] << endl;
        break;
      }
    }
  }
  return 0;
}

```