#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.141592653589793238462643383279;
int n, k, i, j, tree[2000005], m, t[200005], ans[200005];
struct str {
  int x, a, id;
} a[200005], q[200005];
bool cmp(str a, str b) {
  if (a.a == b.a) return a.x < b.x;
  return a.a > b.a;
}
bool cmp2(str a, str b) { return a.a < b.a; }
void modify(int i, int l, int r, int x) {
  if (l == r) {
    tree[i] = 1;
    return;
  }
  int mid = l + r >> 1;
  if (mid >= x)
    modify(i << 1, l, mid, x);
  else
    modify(i << 1 | 1, mid + 1, r, x);
  tree[i] = tree[i << 1] + tree[i << 1 | 1];
}
int Query(int i, int l, int r, int y) {
  if (l == r) return t[l];
  int mid = l + r >> 1;
  if (y > tree[i << 1])
    return Query(i << 1 | 1, mid + 1, r, y - tree[i << 1]);
  else
    return Query(i << 1, l, mid, y);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &a[i].a);
    t[i] = a[i].a;
    a[i].x = i;
  }
  sort(a + 1, a + 1 + n, cmp);
  scanf("%d", &m);
  for (i = 1; i <= m; ++i) {
    scanf("%d %d", &q[i].a, &q[i].x);
    q[i].id = i;
  }
  sort(q + 1, q + 1 + m, cmp2);
  int l = 1;
  for (i = 1; i <= m; ++i) {
    for (; l <= q[i].a; ++l) modify(1, 1, n, a[l].x);
    ans[q[i].id] = Query(1, 1, n, q[i].x);
  }
  for (i = 1; i <= m; ++i) printf("%d\n", ans[i]);
}

### Prompt

Generate a Cpp solution to 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 long double PI = 3.141592653589793238462643383279;
int n, k, i, j, tree[2000005], m, t[200005], ans[200005];
struct str {
  int x, a, id;
} a[200005], q[200005];
bool cmp(str a, str b) {
  if (a.a == b.a) return a.x < b.x;
  return a.a > b.a;
}
bool cmp2(str a, str b) { return a.a < b.a; }
void modify(int i, int l, int r, int x) {
  if (l == r) {
    tree[i] = 1;
    return;
  }
  int mid = l + r >> 1;
  if (mid >= x)
    modify(i << 1, l, mid, x);
  else
    modify(i << 1 | 1, mid + 1, r, x);
  tree[i] = tree[i << 1] + tree[i << 1 | 1];
}
int Query(int i, int l, int r, int y) {
  if (l == r) return t[l];
  int mid = l + r >> 1;
  if (y > tree[i << 1])
    return Query(i << 1 | 1, mid + 1, r, y - tree[i << 1]);
  else
    return Query(i << 1, l, mid, y);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &a[i].a);
    t[i] = a[i].a;
    a[i].x = i;
  }
  sort(a + 1, a + 1 + n, cmp);
  scanf("%d", &m);
  for (i = 1; i <= m; ++i) {
    scanf("%d %d", &q[i].a, &q[i].x);
    q[i].id = i;
  }
  sort(q + 1, q + 1 + m, cmp2);
  int l = 1;
  for (i = 1; i <= m; ++i) {
    for (; l <= q[i].a; ++l) modify(1, 1, n, a[l].x);
    ans[q[i].id] = Query(1, 1, n, q[i].x);
  }
  for (i = 1; i <= m; ++i) printf("%d\n", ans[i]);
}

```