#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void pv(vector<T> v) {
  for (T x : v) cout << x << " ";
  cout << "\n";
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
const int N = 1e9 + 7;
const int mod = 998244353;
vector<int> BIT(200005, 0), ans, b;
vector<pair<int, int>> a;
vector<pair<pair<int, int>, int>> q;
void add(int i, int x) {
  for (; i < BIT.size(); i += i & (-i)) BIT[i] += x;
}
int get(int i) {
  int sum = 0;
  for (; i > 0; i -= i & (-i)) sum += BIT[i];
  return sum;
}
int bit_search(long long val) {
  long long sum = 0, pos = 0;
  for (int i = log2(BIT.size()); i >= 0; i--)
    if (pos + (1 << i) < BIT.size() && sum + BIT[pos + (1 << i)] < val)
      sum += BIT[pos + (1 << i)], pos += (1 << i);
  return pos + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n = read(), k, pos, m;
  b.push_back(N);
  for (int i = 0; i < (n); i++) b.push_back(read());
  for (int i = 0; i < (n + 1); i++) a.push_back(make_pair(-b[i], i));
  m = read();
  for (int i = 0; i < (m); i++) {
    k = read(), pos = read();
    q.push_back(make_pair(make_pair(k, pos), i));
  }
  sort((a).begin(), (a).end());
  sort((q).begin(), (q).end());
  int ind = 1;
  ans.resize(m, -1);
  for (int i = 0; i < (m); i++) {
    while (q[i].first.first >= ind) add(a[ind].second, 1), ind++;
    pos = bit_search(q[i].first.second);
    ans[q[i].second] = pos;
  }
  for (int x : ans) cout << b[x] << "\n";
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 <typename T>
inline void pv(vector<T> v) {
  for (T x : v) cout << x << " ";
  cout << "\n";
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
const int N = 1e9 + 7;
const int mod = 998244353;
vector<int> BIT(200005, 0), ans, b;
vector<pair<int, int>> a;
vector<pair<pair<int, int>, int>> q;
void add(int i, int x) {
  for (; i < BIT.size(); i += i & (-i)) BIT[i] += x;
}
int get(int i) {
  int sum = 0;
  for (; i > 0; i -= i & (-i)) sum += BIT[i];
  return sum;
}
int bit_search(long long val) {
  long long sum = 0, pos = 0;
  for (int i = log2(BIT.size()); i >= 0; i--)
    if (pos + (1 << i) < BIT.size() && sum + BIT[pos + (1 << i)] < val)
      sum += BIT[pos + (1 << i)], pos += (1 << i);
  return pos + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n = read(), k, pos, m;
  b.push_back(N);
  for (int i = 0; i < (n); i++) b.push_back(read());
  for (int i = 0; i < (n + 1); i++) a.push_back(make_pair(-b[i], i));
  m = read();
  for (int i = 0; i < (m); i++) {
    k = read(), pos = read();
    q.push_back(make_pair(make_pair(k, pos), i));
  }
  sort((a).begin(), (a).end());
  sort((q).begin(), (q).end());
  int ind = 1;
  ans.resize(m, -1);
  for (int i = 0; i < (m); i++) {
    while (q[i].first.first >= ind) add(a[ind].second, 1), ind++;
    pos = bit_search(q[i].first.second);
    ans[q[i].second] = pos;
  }
  for (int x : ans) cout << b[x] << "\n";
  return 0;
}

```