#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 1;
int n, q, b[maxn], c[maxn], ans[maxn];
struct node {
  int num, id;
} a[maxn];
struct cxk {
  int k, pos, id;
} s[maxn];
bool cmp(const node &a, const node &b) {
  return a.num == b.num ? a.id < b.id : a.num > b.num;
}
bool cmp2(const cxk &x, const cxk &y) { return x.k < y.k; }
void update(int x) {
  while (x < maxn) c[x]++, x += ((x) & (-x));
}
int query(int x) {
  int res = 0;
  while (x > 0) res += c[x], x -= ((x) & (-x));
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%d", &a[i].num), b[i] = a[i].num, a[i].id = i;
  sort(a + 1, a + 1 + n, cmp);
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) scanf("%d%d", &s[i].k, &s[i].pos), s[i].id = i;
  sort(s + 1, s + 1 + q, cmp2);
  int k = 1;
  for (int i = 1; i <= q; ++i) {
    while (k <= s[i].k) update(a[k].id), k++;
    int l = 1, r = n;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (query(mid) >= s[i].pos)
        r = mid;
      else
        l = mid + 1;
    }
    ans[s[i].id] = b[l];
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
}

### 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;
const int maxn = 2e5 + 1;
int n, q, b[maxn], c[maxn], ans[maxn];
struct node {
  int num, id;
} a[maxn];
struct cxk {
  int k, pos, id;
} s[maxn];
bool cmp(const node &a, const node &b) {
  return a.num == b.num ? a.id < b.id : a.num > b.num;
}
bool cmp2(const cxk &x, const cxk &y) { return x.k < y.k; }
void update(int x) {
  while (x < maxn) c[x]++, x += ((x) & (-x));
}
int query(int x) {
  int res = 0;
  while (x > 0) res += c[x], x -= ((x) & (-x));
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%d", &a[i].num), b[i] = a[i].num, a[i].id = i;
  sort(a + 1, a + 1 + n, cmp);
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) scanf("%d%d", &s[i].k, &s[i].pos), s[i].id = i;
  sort(s + 1, s + 1 + q, cmp2);
  int k = 1;
  for (int i = 1; i <= q; ++i) {
    while (k <= s[i].k) update(a[k].id), k++;
    int l = 1, r = n;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (query(mid) >= s[i].pos)
        r = mid;
      else
        l = mid + 1;
    }
    ans[s[i].id] = b[l];
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
}

```