#include <bits/stdc++.h>
using namespace std;
clock_t timeStart, timeFinish;
void timeBegin() { timeStart = clock(); }
void timeEnd() { timeFinish = clock(); }
void timeDuration() {
  timeEnd();
  double time_taken = double(timeFinish - timeStart) / double(CLOCKS_PER_SEC);
  cout << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cout << " sec " << endl;
}
class InputReader {
 public:
  InputReader() {
    input_file = stdin;
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  InputReader(const char *file_name) {
    input_file = fopen(file_name, "r");
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  inline InputReader &operator>>(int &n) {
    while ((buffer[cursor] < '0' || buffer[cursor] > '9') &&
           buffer[cursor] != '-') {
      advance();
    }
    int sign = 1;
    if (buffer[cursor] == '-') {
      sign = -1;
      advance();
    }
    n = 0;
    while ('0' <= buffer[cursor] && buffer[cursor] <= '9') {
      n = n * 10 + buffer[cursor] - '0';
      advance();
    }
    n *= sign;
    return *this;
  }

 private:
  FILE *input_file;
  static const int SIZE = 1 << 17;
  int cursor;
  char buffer[SIZE];
  inline void advance() {
    ++cursor;
    if (cursor == SIZE) {
      cursor = 0;
      fread(buffer, SIZE, 1, input_file);
    }
  }
};
const int MAXN = 200000;
const int MAXLOG = 20;
int val[1 + MAXN];
int v[1 + MAXN];
void normalize(int n, int &m) {
  map<int, int> code;
  for (int i = 1; i <= n; i++) {
    code[v[i]] = 1;
  }
  for (auto &it : code) {
    it.second = ++m;
    val[it.second] = it.first;
  }
  for (int i = 1; i <= n; i++) {
    v[i] = code[v[i]];
  }
}
int fenwick[1 + MAXN], answer[1 + MAXN];
vector<pair<int, int> > queries[1 + MAXN];
pair<int, int> vn[1 + MAXN];
void update(int x, int add, int n) {
  for (; x <= n; x += (x & -x)) {
    fenwick[x] += add;
  }
}
int getPos(int sum, int n) {
  int answer = 0;
  for (int step = MAXLOG; step >= 0; step--) {
    if (answer + (1 << step) <= n && fenwick[answer + (1 << step)] <= sum) {
      sum -= fenwick[answer + (1 << step)];
      answer += (1 << step);
    }
  }
  return answer;
}
int main() {
  timeBegin();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  int m = 0;
  normalize(n, m);
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int k, p;
    cin >> k >> p;
    queries[k].push_back({p, i});
  }
  for (int i = 1; i <= n; i++) {
    update(i, 1, n);
    vn[i] = {v[i], -i};
  }
  sort(vn + 1, vn + n + 1);
  for (int i = 1; i <= n; i++) {
    int k = n - i + 1;
    for (auto it : queries[k]) {
      int p = it.first, id = it.second;
      answer[id] = getPos(p - 1, n) + 1;
    }
    update(-vn[i].second, -1, n);
  }
  for (int i = 1; i <= q; i++) {
    cout << val[v[answer[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;
clock_t timeStart, timeFinish;
void timeBegin() { timeStart = clock(); }
void timeEnd() { timeFinish = clock(); }
void timeDuration() {
  timeEnd();
  double time_taken = double(timeFinish - timeStart) / double(CLOCKS_PER_SEC);
  cout << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cout << " sec " << endl;
}
class InputReader {
 public:
  InputReader() {
    input_file = stdin;
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  InputReader(const char *file_name) {
    input_file = fopen(file_name, "r");
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  inline InputReader &operator>>(int &n) {
    while ((buffer[cursor] < '0' || buffer[cursor] > '9') &&
           buffer[cursor] != '-') {
      advance();
    }
    int sign = 1;
    if (buffer[cursor] == '-') {
      sign = -1;
      advance();
    }
    n = 0;
    while ('0' <= buffer[cursor] && buffer[cursor] <= '9') {
      n = n * 10 + buffer[cursor] - '0';
      advance();
    }
    n *= sign;
    return *this;
  }

 private:
  FILE *input_file;
  static const int SIZE = 1 << 17;
  int cursor;
  char buffer[SIZE];
  inline void advance() {
    ++cursor;
    if (cursor == SIZE) {
      cursor = 0;
      fread(buffer, SIZE, 1, input_file);
    }
  }
};
const int MAXN = 200000;
const int MAXLOG = 20;
int val[1 + MAXN];
int v[1 + MAXN];
void normalize(int n, int &m) {
  map<int, int> code;
  for (int i = 1; i <= n; i++) {
    code[v[i]] = 1;
  }
  for (auto &it : code) {
    it.second = ++m;
    val[it.second] = it.first;
  }
  for (int i = 1; i <= n; i++) {
    v[i] = code[v[i]];
  }
}
int fenwick[1 + MAXN], answer[1 + MAXN];
vector<pair<int, int> > queries[1 + MAXN];
pair<int, int> vn[1 + MAXN];
void update(int x, int add, int n) {
  for (; x <= n; x += (x & -x)) {
    fenwick[x] += add;
  }
}
int getPos(int sum, int n) {
  int answer = 0;
  for (int step = MAXLOG; step >= 0; step--) {
    if (answer + (1 << step) <= n && fenwick[answer + (1 << step)] <= sum) {
      sum -= fenwick[answer + (1 << step)];
      answer += (1 << step);
    }
  }
  return answer;
}
int main() {
  timeBegin();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  int m = 0;
  normalize(n, m);
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int k, p;
    cin >> k >> p;
    queries[k].push_back({p, i});
  }
  for (int i = 1; i <= n; i++) {
    update(i, 1, n);
    vn[i] = {v[i], -i};
  }
  sort(vn + 1, vn + n + 1);
  for (int i = 1; i <= n; i++) {
    int k = n - i + 1;
    for (auto it : queries[k]) {
      int p = it.first, id = it.second;
      answer[id] = getPos(p - 1, n) + 1;
    }
    update(-vn[i].second, -1, n);
  }
  for (int i = 1; i <= q; i++) {
    cout << val[v[answer[i]]] << "\n";
  }
  return 0;
}

```