#include <bits/stdc++.h>
using namespace std;
int A[200005], F[200005], Y[200005];
int BIT[200005];
map<int, int> H;
vector<int> X[200005];
vector<vector<int> > queries;
void update(int i) {
  for (; i <= 200000; i += (i & (-i))) BIT[i]++;
}
int query(int i) {
  int ans = 0;
  for (; i > 0; i -= (i & (-i))) ans += BIT[i];
  return ans;
}
int solve(int ptr, int k, int done, int pos, int N) {
  int left = 1, right = N, ans = 1;
  while (left <= right) {
    int mid = (left + right) / 2;
    int curr = upper_bound(X[ptr].begin(), X[ptr].end(), mid) - X[ptr].begin();
    curr = min(curr, k - done) + query(mid);
    if (curr >= pos)
      ans = mid, right = mid - 1;
    else
      left = mid + 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  int N;
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    H[A[i]];
  }
  int h = 0;
  for (auto &it : H) it.second = ++h;
  for (int i = 1; i <= N; i++) {
    int h = H[A[i]];
    Y[h] = A[i];
    X[h].push_back(i);
  }
  int M;
  cin >> M;
  for (int i = 1; i <= M; i++) {
    int k, pos;
    cin >> k >> pos;
    queries.push_back({k, pos, i});
  }
  int ptr = h, done = 0;
  sort(queries.begin(), queries.end());
  for (auto q : queries) {
    int k = q[0], pos = q[1], indx = q[2];
    while (k > done + X[ptr].size()) {
      for (auto x : X[ptr]) update(x);
      done += X[ptr--].size();
    }
    F[indx] = A[solve(ptr, k, done, pos, N)];
  }
  for (int i = 1; i <= M; i++) cout << F[i] << "\n";
  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;
int A[200005], F[200005], Y[200005];
int BIT[200005];
map<int, int> H;
vector<int> X[200005];
vector<vector<int> > queries;
void update(int i) {
  for (; i <= 200000; i += (i & (-i))) BIT[i]++;
}
int query(int i) {
  int ans = 0;
  for (; i > 0; i -= (i & (-i))) ans += BIT[i];
  return ans;
}
int solve(int ptr, int k, int done, int pos, int N) {
  int left = 1, right = N, ans = 1;
  while (left <= right) {
    int mid = (left + right) / 2;
    int curr = upper_bound(X[ptr].begin(), X[ptr].end(), mid) - X[ptr].begin();
    curr = min(curr, k - done) + query(mid);
    if (curr >= pos)
      ans = mid, right = mid - 1;
    else
      left = mid + 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  int N;
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    H[A[i]];
  }
  int h = 0;
  for (auto &it : H) it.second = ++h;
  for (int i = 1; i <= N; i++) {
    int h = H[A[i]];
    Y[h] = A[i];
    X[h].push_back(i);
  }
  int M;
  cin >> M;
  for (int i = 1; i <= M; i++) {
    int k, pos;
    cin >> k >> pos;
    queries.push_back({k, pos, i});
  }
  int ptr = h, done = 0;
  sort(queries.begin(), queries.end());
  for (auto q : queries) {
    int k = q[0], pos = q[1], indx = q[2];
    while (k > done + X[ptr].size()) {
      for (auto x : X[ptr]) update(x);
      done += X[ptr--].size();
    }
    F[indx] = A[solve(ptr, k, done, pos, N)];
  }
  for (int i = 1; i <= M; i++) cout << F[i] << "\n";
  return 0;
}

```