#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int NMAX = 1123456;
int in[NMAX];
pair<int, int> toSort[NMAX];
int inter[NMAX];
int interstart = 0;
int n, m;
struct Ask {
  int kolik, kam, i;
} ask[NMAX];
int out[NMAX];
void interAdd(int i, int val) {
  if (0) printf("ADD %d %d\n", i, val);
  i += interstart;
  inter[i] = val;
  for (i >>= 1; i; i >>= 1) inter[i] = inter[2 * i] + inter[2 * i + 1];
}
int interIndexPrefix(int p) {
  if (0) printf("PREFIX %d", p);
  int i;
  for (i = 1;; i *= 2) {
    if (inter[i] < p) {
      p -= inter[i];
      i++;
    }
    if (i >= interstart) {
      if (0) printf(" -> %d\n", i - interstart);
      return i - interstart;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", in + i);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &(ask[i].kolik), &(ask[i].kam));
    ask[i].i = i;
  }
  for (int i = 0; i < n; i++) {
    toSort[i] = {-in[i], i};
  }
  for (interstart = 1; interstart <= n; interstart *= 2)
    ;
  sort(toSort, toSort + n);
  sort(ask, ask + m, [](Ask a, Ask b) -> bool { return a.kolik < b.kolik; });
  int last = 0;
  for (int i = 0; i < m; i++) {
    if (0) printf("--%d %d[%d] %d\n", ask[i].i, ask[i].kolik, last, ask[i].kam);
    for (int j = last; j < ask[i].kolik; j++) interAdd(toSort[j].second, 1);
    last = ask[i].kolik;
    out[ask[i].i] = in[interIndexPrefix(ask[i].kam)];
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", out[i]);
  }
  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;
using ll = long long;
const int NMAX = 1123456;
int in[NMAX];
pair<int, int> toSort[NMAX];
int inter[NMAX];
int interstart = 0;
int n, m;
struct Ask {
  int kolik, kam, i;
} ask[NMAX];
int out[NMAX];
void interAdd(int i, int val) {
  if (0) printf("ADD %d %d\n", i, val);
  i += interstart;
  inter[i] = val;
  for (i >>= 1; i; i >>= 1) inter[i] = inter[2 * i] + inter[2 * i + 1];
}
int interIndexPrefix(int p) {
  if (0) printf("PREFIX %d", p);
  int i;
  for (i = 1;; i *= 2) {
    if (inter[i] < p) {
      p -= inter[i];
      i++;
    }
    if (i >= interstart) {
      if (0) printf(" -> %d\n", i - interstart);
      return i - interstart;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", in + i);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &(ask[i].kolik), &(ask[i].kam));
    ask[i].i = i;
  }
  for (int i = 0; i < n; i++) {
    toSort[i] = {-in[i], i};
  }
  for (interstart = 1; interstart <= n; interstart *= 2)
    ;
  sort(toSort, toSort + n);
  sort(ask, ask + m, [](Ask a, Ask b) -> bool { return a.kolik < b.kolik; });
  int last = 0;
  for (int i = 0; i < m; i++) {
    if (0) printf("--%d %d[%d] %d\n", ask[i].i, ask[i].kolik, last, ask[i].kam);
    for (int j = last; j < ask[i].kolik; j++) interAdd(toSort[j].second, 1);
    last = ask[i].kolik;
    out[ask[i].i] = in[interIndexPrefix(ask[i].kam)];
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", out[i]);
  }
  return 0;
}

```