#include <bits/stdc++.h>
using namespace std;
inline void read(int &first) {
  register int c = getchar();
  first = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    first = (first << 1) + (first << 3) + c - 48;
  }
  if (neg) first = -first;
}
inline void read(long long &first) {
  register int c = getchar();
  first = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    first = (first << 1) + (first << 3) + c - 48;
  }
  if (neg) first = -first;
}
inline void writeln(long long first) {
  char buffor[21];
  register int i = 0;
  int neg = 0;
  if (first < 0) {
    neg = 1;
    first = -first;
  }
  do {
    buffor[i++] = (first % 10) + '0';
    first /= 10;
  } while (first);
  i--;
  if (neg) putchar('-');
  while (i >= 0) putchar(buffor[i--]);
  putchar('\n');
}
inline void write(long long first) {
  char buffor[21];
  register int i = 0;
  int neg = 0;
  if (first < 0) {
    neg = 1;
    first = -first;
  }
  do {
    buffor[i++] = (first % 10) + '0';
    first /= 10;
  } while (first);
  i--;
  if (neg) putchar('-');
  while (i >= 0) putchar(buffor[i--]);
  putchar(' ');
}
const int N = 2e5 + 7;
int n, m;
int b[N], ans[N];
pair<int, int> a[N];
vector<pair<int, int> > query[N];
struct bit {
  int tree[N];
  void upd(int first) {
    for (; first < N; first += first & (-first)) tree[first]++;
  }
  int get(int first) {
    int res = 0;
    for (; first > 0; first -= first & (-first)) res += tree[first];
    return res;
  }
} BIT;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  if (fopen("test.inp", "r"))
    freopen("test.inp", "r", stdin), freopen("test.out", "w", stdout);
  read(n);
  for (int i = 1; i <= n; i++) {
    read(a[i].first);
    a[i].second = i;
    b[i] = a[i].first;
  }
  sort(a + 1, a + n + 1, [](pair<int, int> a, pair<int, int> b) {
    return a.first > b.first || (a.first == b.first && (a.second < b.second));
  });
  read(m);
  for (int i = 1; i <= m; i++) {
    int k, pos;
    read(k);
    read(pos);
    query[k].push_back(pair<int, int>(pos, i));
  }
  for (int i = 1; i <= n; i++) {
    BIT.upd(a[i].second);
    for (auto t : query[i]) {
      int first = t.first;
      int id = t.second;
      int l = 1;
      int r = n;
      while (l <= r) {
        int mid = l + r >> 1;
        if (BIT.get(mid) < first)
          l = mid + 1;
        else
          r = mid - 1;
      }
      ans[id] = b[l];
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
}

### Prompt

Please provide a Cpp coded solution 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;
inline void read(int &first) {
  register int c = getchar();
  first = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    first = (first << 1) + (first << 3) + c - 48;
  }
  if (neg) first = -first;
}
inline void read(long long &first) {
  register int c = getchar();
  first = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    first = (first << 1) + (first << 3) + c - 48;
  }
  if (neg) first = -first;
}
inline void writeln(long long first) {
  char buffor[21];
  register int i = 0;
  int neg = 0;
  if (first < 0) {
    neg = 1;
    first = -first;
  }
  do {
    buffor[i++] = (first % 10) + '0';
    first /= 10;
  } while (first);
  i--;
  if (neg) putchar('-');
  while (i >= 0) putchar(buffor[i--]);
  putchar('\n');
}
inline void write(long long first) {
  char buffor[21];
  register int i = 0;
  int neg = 0;
  if (first < 0) {
    neg = 1;
    first = -first;
  }
  do {
    buffor[i++] = (first % 10) + '0';
    first /= 10;
  } while (first);
  i--;
  if (neg) putchar('-');
  while (i >= 0) putchar(buffor[i--]);
  putchar(' ');
}
const int N = 2e5 + 7;
int n, m;
int b[N], ans[N];
pair<int, int> a[N];
vector<pair<int, int> > query[N];
struct bit {
  int tree[N];
  void upd(int first) {
    for (; first < N; first += first & (-first)) tree[first]++;
  }
  int get(int first) {
    int res = 0;
    for (; first > 0; first -= first & (-first)) res += tree[first];
    return res;
  }
} BIT;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  if (fopen("test.inp", "r"))
    freopen("test.inp", "r", stdin), freopen("test.out", "w", stdout);
  read(n);
  for (int i = 1; i <= n; i++) {
    read(a[i].first);
    a[i].second = i;
    b[i] = a[i].first;
  }
  sort(a + 1, a + n + 1, [](pair<int, int> a, pair<int, int> b) {
    return a.first > b.first || (a.first == b.first && (a.second < b.second));
  });
  read(m);
  for (int i = 1; i <= m; i++) {
    int k, pos;
    read(k);
    read(pos);
    query[k].push_back(pair<int, int>(pos, i));
  }
  for (int i = 1; i <= n; i++) {
    BIT.upd(a[i].second);
    for (auto t : query[i]) {
      int first = t.first;
      int id = t.second;
      int l = 1;
      int r = n;
      while (l <= r) {
        int mid = l + r >> 1;
        if (BIT.get(mid) < first)
          l = mid + 1;
        else
          r = mid - 1;
      }
      ans[id] = b[l];
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
}

```