#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 100;
struct node {
  int id;
  int num;
};
node s[maxn];
int b[maxn];
int num[maxn];
int N;
bool cmp(node a, node b) {
  if (a.num != b.num)
    return a.num > b.num;
  else
    return a.id < b.id;
}
struct ask {
  int id;
  int L = 0;
  int pos;
};
ask A[maxn];
bool cmp2(ask a, ask b) { return a.L < b.L; }
int ans[maxn];
int C[maxn];
void add(int x) {
  while (x < maxn) {
    C[x] += 1;
    x += (x) & (-x);
  }
}
int getsum(int x) {
  int sum = 0;
  while (x > 0) {
    sum += C[x];
    x -= (x) & (-x);
  }
  return sum;
}
int f(int pos) {
  int L = 1;
  int R = N;
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (getsum(mid) < pos)
      L = mid + 1;
    else
      R = mid - 1;
  }
  return L;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> num[i];
  for (int i = 1; i <= N; i++) s[i].num = num[i], s[i].id = i;
  sort(s + 1, s + 1 + N, cmp);
  for (int i = 1; i <= N; i++) b[i] = s[i].id;
  int Q;
  cin >> Q;
  for (int i = 1; i <= Q; i++) {
    cin >> A[i].L;
    A[i].id = i;
    cin >> A[i].pos;
  }
  sort(A + 1, A + 1 + Q, cmp2);
  int L = 1;
  for (int i = 1; i <= Q; i++) {
    for (; L <= A[i].L; L++) add(b[L]);
    ans[A[i].id] = f(A[i].pos);
  }
  for (int i = 1; i <= Q; i++) cout << num[ans[i]] << endl;
  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;
const int maxn = 2e5 + 100;
struct node {
  int id;
  int num;
};
node s[maxn];
int b[maxn];
int num[maxn];
int N;
bool cmp(node a, node b) {
  if (a.num != b.num)
    return a.num > b.num;
  else
    return a.id < b.id;
}
struct ask {
  int id;
  int L = 0;
  int pos;
};
ask A[maxn];
bool cmp2(ask a, ask b) { return a.L < b.L; }
int ans[maxn];
int C[maxn];
void add(int x) {
  while (x < maxn) {
    C[x] += 1;
    x += (x) & (-x);
  }
}
int getsum(int x) {
  int sum = 0;
  while (x > 0) {
    sum += C[x];
    x -= (x) & (-x);
  }
  return sum;
}
int f(int pos) {
  int L = 1;
  int R = N;
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (getsum(mid) < pos)
      L = mid + 1;
    else
      R = mid - 1;
  }
  return L;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> num[i];
  for (int i = 1; i <= N; i++) s[i].num = num[i], s[i].id = i;
  sort(s + 1, s + 1 + N, cmp);
  for (int i = 1; i <= N; i++) b[i] = s[i].id;
  int Q;
  cin >> Q;
  for (int i = 1; i <= Q; i++) {
    cin >> A[i].L;
    A[i].id = i;
    cin >> A[i].pos;
  }
  sort(A + 1, A + 1 + Q, cmp2);
  int L = 1;
  for (int i = 1; i <= Q; i++) {
    for (; L <= A[i].L; L++) add(b[L]);
    ans[A[i].id] = f(A[i].pos);
  }
  for (int i = 1; i <= Q; i++) cout << num[ans[i]] << endl;
  return 0;
}

```