#include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
class BIT {
 public:
  vector<int> bit;
  int size_ = 0;
  void init(int sz) {
    size_ = sz + 2;
    bit.resize(size_ + 2, 0);
  }
  void add(int pos, int x) {
    pos++;
    while (pos <= size_) {
      bit[pos] += x;
      pos += (pos & -pos);
    }
  }
  int sum(int pos) {
    int s = 0;
    pos++;
    while (pos >= 1) {
      s += bit[pos];
      pos -= (pos & -pos);
    }
    return s;
  }
  int searchs(int pos) {
    int cl = 0, cr = size_ + 1, cm, minx = (1 << 30);
    for (int i = 0; i < 22; i++) {
      cm = ((cl + cr) >> 1);
      int v = sum(cm);
      if (v >= pos) {
        minx = min(minx, cm);
        cr = cm;
      } else {
        cl = cm;
      }
    }
    return minx;
  }
};
int N, A[1 << 18];
int Q, K[1 << 18], P[1 << 18], ans[1 << 18];
vector<pair<int, int>> rem;
vector<int> G[1 << 18];
BIT Z;
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) scanf("%d%d", &K[i], &P[i]);
  for (int i = 1; i <= N; i++) rem.push_back(make_pair(-A[i], i));
  for (int i = 1; i <= Q; i++) G[K[i]].push_back(i);
  sort(rem.begin(), rem.end());
  Z.init(N + 2);
  for (int i = 1; i <= N; i++) {
    Z.add(rem[i - 1].second, 1);
    for (int j = 0; j < G[i].size(); j++) {
      int idx = G[i][j];
      int val = Z.searchs(P[idx]);
      ans[idx] = A[val];
    }
  }
  for (int i = 1; i <= Q; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Please formulate 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;
#pragma warning(disable : 4996)
class BIT {
 public:
  vector<int> bit;
  int size_ = 0;
  void init(int sz) {
    size_ = sz + 2;
    bit.resize(size_ + 2, 0);
  }
  void add(int pos, int x) {
    pos++;
    while (pos <= size_) {
      bit[pos] += x;
      pos += (pos & -pos);
    }
  }
  int sum(int pos) {
    int s = 0;
    pos++;
    while (pos >= 1) {
      s += bit[pos];
      pos -= (pos & -pos);
    }
    return s;
  }
  int searchs(int pos) {
    int cl = 0, cr = size_ + 1, cm, minx = (1 << 30);
    for (int i = 0; i < 22; i++) {
      cm = ((cl + cr) >> 1);
      int v = sum(cm);
      if (v >= pos) {
        minx = min(minx, cm);
        cr = cm;
      } else {
        cl = cm;
      }
    }
    return minx;
  }
};
int N, A[1 << 18];
int Q, K[1 << 18], P[1 << 18], ans[1 << 18];
vector<pair<int, int>> rem;
vector<int> G[1 << 18];
BIT Z;
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) scanf("%d%d", &K[i], &P[i]);
  for (int i = 1; i <= N; i++) rem.push_back(make_pair(-A[i], i));
  for (int i = 1; i <= Q; i++) G[K[i]].push_back(i);
  sort(rem.begin(), rem.end());
  Z.init(N + 2);
  for (int i = 1; i <= N; i++) {
    Z.add(rem[i - 1].second, 1);
    for (int j = 0; j < G[i].size(); j++) {
      int idx = G[i][j];
      int val = Z.searchs(P[idx]);
      ans[idx] = A[val];
    }
  }
  for (int i = 1; i <= Q; i++) printf("%d\n", ans[i]);
  return 0;
}

```