#include <bits/stdc++.h>
using namespace std;
int n, m, i, j, a[200005], ans[200005], pos[200005], sum[800005];
vector<int> qry[200005];
struct ii {
  int x, id;
} b[200005];
bool cmp(ii x, ii y) {
  if (x.x != y.x) return x.x > y.x;
  return x.id < y.id;
}
void update(int x, int l, int r, int y, int c) {
  sum[x] += c;
  if (l == r) return;
  int mid = (l + r) / 2;
  if (y <= mid)
    update(x + x, l, mid, y, c);
  else
    update(x + x + 1, mid + 1, r, y, c);
}
int query(int x, int l, int r, int y) {
  if (l == r) {
    return l;
  }
  int mid = (l + r) / 2;
  if (sum[x + x] >= y) return query(x + x, l, mid, y);
  return query(x + x + 1, mid + 1, r, y - sum[x + x]);
}
int main() {
  scanf("%d", &n);
  for ((i) = 1; (i) <= (n); ++(i)) {
    scanf("%d", &a[i]);
    b[i] = (ii){a[i], i};
  }
  sort(b + 1, b + n + 1, cmp);
  scanf("%d", &m);
  for ((i) = 1; (i) <= (m); ++(i)) {
    int x;
    scanf("%d%d", &x, &pos[i]);
    qry[x].push_back(i);
  }
  for ((i) = 1; (i) <= (n); ++(i)) {
    update(1, 1, n, b[i].id, 1);
    for (__typeof((qry[i]).begin()) it = (qry[i]).begin(); it != (qry[i]).end();
         it++) {
      ans[*it] = a[query(1, 1, n, pos[*it])];
    }
  }
  for ((i) = 1; (i) <= (m); ++(i)) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 n, m, i, j, a[200005], ans[200005], pos[200005], sum[800005];
vector<int> qry[200005];
struct ii {
  int x, id;
} b[200005];
bool cmp(ii x, ii y) {
  if (x.x != y.x) return x.x > y.x;
  return x.id < y.id;
}
void update(int x, int l, int r, int y, int c) {
  sum[x] += c;
  if (l == r) return;
  int mid = (l + r) / 2;
  if (y <= mid)
    update(x + x, l, mid, y, c);
  else
    update(x + x + 1, mid + 1, r, y, c);
}
int query(int x, int l, int r, int y) {
  if (l == r) {
    return l;
  }
  int mid = (l + r) / 2;
  if (sum[x + x] >= y) return query(x + x, l, mid, y);
  return query(x + x + 1, mid + 1, r, y - sum[x + x]);
}
int main() {
  scanf("%d", &n);
  for ((i) = 1; (i) <= (n); ++(i)) {
    scanf("%d", &a[i]);
    b[i] = (ii){a[i], i};
  }
  sort(b + 1, b + n + 1, cmp);
  scanf("%d", &m);
  for ((i) = 1; (i) <= (m); ++(i)) {
    int x;
    scanf("%d%d", &x, &pos[i]);
    qry[x].push_back(i);
  }
  for ((i) = 1; (i) <= (n); ++(i)) {
    update(1, 1, n, b[i].id, 1);
    for (__typeof((qry[i]).begin()) it = (qry[i]).begin(); it != (qry[i]).end();
         it++) {
      ans[*it] = a[query(1, 1, n, pos[*it])];
    }
  }
  for ((i) = 1; (i) <= (m); ++(i)) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```