#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 2e5 + 200;
int n, ai[MAX_N], q, nodes[MAX_N << 2], ans[MAX_N];
struct query {
  int len, pos, id;
  bool operator<(const query &rhs) const { return len < rhs.len; }
} queries[MAX_N];
void build(int l, int r, int p) {
  if (l == r) return (void)(nodes[p] = 1);
  build(l, ((l + r) >> 1), (p << 1)),
      build(((l + r) >> 1) + 1, r, ((p << 1) | 1));
  nodes[p] = nodes[(p << 1)] + nodes[((p << 1) | 1)];
}
void update(int qx, int l, int r, int p, int val) {
  if (l == r) return (void)(nodes[p] += val);
  if (qx <= ((l + r) >> 1))
    update(qx, l, ((l + r) >> 1), (p << 1), val);
  else
    update(qx, ((l + r) >> 1) + 1, r, ((p << 1) | 1), val);
  nodes[p] = nodes[(p << 1)] + nodes[((p << 1) | 1)];
}
int queryPos(int k, int l, int r, int p) {
  if (l == r) return l;
  int lsiz = nodes[(p << 1)];
  if (k <= lsiz)
    return queryPos(k, l, ((l + r) >> 1), (p << 1));
  else
    return queryPos(k - lsiz, ((l + r) >> 1) + 1, r, ((p << 1) | 1));
}
struct tup {
  int x, y;
  bool operator<(const tup &rhs) const {
    return x < rhs.x || (x == rhs.x && y > rhs.y);
  }
};
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &ai[i]);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++)
    scanf("%d%d", &queries[i].len, &queries[i].pos), queries[i].id = i;
  sort(queries + 1, queries + 1 + q), reverse(queries + 1, queries + 1 + q);
  multiset<tup> ms;
  for (int i = 1; i <= n; i++) ms.insert(tup{ai[i], i});
  build(1, n, 1);
  for (int curt_len = n, qptr = 1; curt_len >= 1; curt_len--) {
    while (qptr <= q && queries[qptr].len == curt_len) {
      int pos = queryPos(queries[qptr].pos, 1, n, 1);
      ans[queries[qptr].id] = ai[pos], qptr++;
    }
    tup del_element = *ms.begin();
    ms.erase(ms.begin());
    update(del_element.y, 1, n, 1, -1);
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[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;
const int MAX_N = 2e5 + 200;
int n, ai[MAX_N], q, nodes[MAX_N << 2], ans[MAX_N];
struct query {
  int len, pos, id;
  bool operator<(const query &rhs) const { return len < rhs.len; }
} queries[MAX_N];
void build(int l, int r, int p) {
  if (l == r) return (void)(nodes[p] = 1);
  build(l, ((l + r) >> 1), (p << 1)),
      build(((l + r) >> 1) + 1, r, ((p << 1) | 1));
  nodes[p] = nodes[(p << 1)] + nodes[((p << 1) | 1)];
}
void update(int qx, int l, int r, int p, int val) {
  if (l == r) return (void)(nodes[p] += val);
  if (qx <= ((l + r) >> 1))
    update(qx, l, ((l + r) >> 1), (p << 1), val);
  else
    update(qx, ((l + r) >> 1) + 1, r, ((p << 1) | 1), val);
  nodes[p] = nodes[(p << 1)] + nodes[((p << 1) | 1)];
}
int queryPos(int k, int l, int r, int p) {
  if (l == r) return l;
  int lsiz = nodes[(p << 1)];
  if (k <= lsiz)
    return queryPos(k, l, ((l + r) >> 1), (p << 1));
  else
    return queryPos(k - lsiz, ((l + r) >> 1) + 1, r, ((p << 1) | 1));
}
struct tup {
  int x, y;
  bool operator<(const tup &rhs) const {
    return x < rhs.x || (x == rhs.x && y > rhs.y);
  }
};
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &ai[i]);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++)
    scanf("%d%d", &queries[i].len, &queries[i].pos), queries[i].id = i;
  sort(queries + 1, queries + 1 + q), reverse(queries + 1, queries + 1 + q);
  multiset<tup> ms;
  for (int i = 1; i <= n; i++) ms.insert(tup{ai[i], i});
  build(1, n, 1);
  for (int curt_len = n, qptr = 1; curt_len >= 1; curt_len--) {
    while (qptr <= q && queries[qptr].len == curt_len) {
      int pos = queryPos(queries[qptr].pos, 1, n, 1);
      ans[queries[qptr].id] = ai[pos], qptr++;
    }
    tup del_element = *ms.begin();
    ms.erase(ms.begin());
    update(del_element.y, 1, n, 1, -1);
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}

```