#include <bits/stdc++.h>
using namespace std;
const int maxx = 1000010;
struct node {
  int num;
  long long val;
  bool operator<(const node &X) const {
    if (val == X.val) return num > X.num;
    return val < X.val;
  }
};
struct nod {
  int num;
  int pos;
  int bj;
} s[maxx];
struct no {
  int l;
  int r;
  int sum;
  long long val;
} tree[maxx];
bool cmp(nod a, nod b) { return a.num < b.num; }
priority_queue<node> q;
long long n, m, k, p;
long long ans = 0, total = 0, sum = 0;
long long a[maxx], b[maxx];
void build(int p, int l, int r) {
  tree[p].l = l, tree[p].r = r;
  int mid = (l + r) >> 1;
  if (l == r) return;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
}
void update(int p, int c, long long val) {
  int l = tree[p].l, r = tree[p].r;
  if (l == r && l == c) {
    tree[p].sum = 1;
    tree[p].val = val;
    return;
  }
  int mid = (l + r) >> 1;
  if (c <= mid)
    update(p << 1, c, val);
  else
    update(p << 1 | 1, c, val);
  tree[p].val = tree[p << 1].val + tree[p << 1 | 1].val;
  tree[p].sum = tree[p << 1].sum + tree[p << 1 | 1].sum;
}
long long query(int p, int rank) {
  int l = tree[p].l, r = tree[p].r;
  int mid = (l + r) >> 1;
  if (l == r) return tree[p].val;
  if (tree[p << 1].sum >= rank)
    return query(p << 1, rank);
  else
    return query(p << 1 | 1, rank - tree[p << 1].sum);
}
int main() {
  scanf("%lld", &n);
  for (register int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    q.push(node{i, a[i]});
  }
  scanf("%lld", &m);
  for (register int i = 1; i <= m; i++) {
    scanf("%d%d", &s[i].num, &s[i].pos);
    s[i].bj = i;
  }
  sort(s + 1, s + m + 1, cmp);
  build(1, 1, n);
  for (register int i = 1; i <= m; i++) {
    while (total < s[i].num) {
      int u = q.top().num;
      long long val = q.top().val;
      update(1, u, val);
      q.pop();
      total++;
    }
    b[s[i].bj] = query(1, s[i].pos);
  }
  for (register int i = 1; i <= m; i++) printf("%lld\n", b[i]);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 maxx = 1000010;
struct node {
  int num;
  long long val;
  bool operator<(const node &X) const {
    if (val == X.val) return num > X.num;
    return val < X.val;
  }
};
struct nod {
  int num;
  int pos;
  int bj;
} s[maxx];
struct no {
  int l;
  int r;
  int sum;
  long long val;
} tree[maxx];
bool cmp(nod a, nod b) { return a.num < b.num; }
priority_queue<node> q;
long long n, m, k, p;
long long ans = 0, total = 0, sum = 0;
long long a[maxx], b[maxx];
void build(int p, int l, int r) {
  tree[p].l = l, tree[p].r = r;
  int mid = (l + r) >> 1;
  if (l == r) return;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
}
void update(int p, int c, long long val) {
  int l = tree[p].l, r = tree[p].r;
  if (l == r && l == c) {
    tree[p].sum = 1;
    tree[p].val = val;
    return;
  }
  int mid = (l + r) >> 1;
  if (c <= mid)
    update(p << 1, c, val);
  else
    update(p << 1 | 1, c, val);
  tree[p].val = tree[p << 1].val + tree[p << 1 | 1].val;
  tree[p].sum = tree[p << 1].sum + tree[p << 1 | 1].sum;
}
long long query(int p, int rank) {
  int l = tree[p].l, r = tree[p].r;
  int mid = (l + r) >> 1;
  if (l == r) return tree[p].val;
  if (tree[p << 1].sum >= rank)
    return query(p << 1, rank);
  else
    return query(p << 1 | 1, rank - tree[p << 1].sum);
}
int main() {
  scanf("%lld", &n);
  for (register int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    q.push(node{i, a[i]});
  }
  scanf("%lld", &m);
  for (register int i = 1; i <= m; i++) {
    scanf("%d%d", &s[i].num, &s[i].pos);
    s[i].bj = i;
  }
  sort(s + 1, s + m + 1, cmp);
  build(1, 1, n);
  for (register int i = 1; i <= m; i++) {
    while (total < s[i].num) {
      int u = q.top().num;
      long long val = q.top().val;
      update(1, u, val);
      q.pop();
      total++;
    }
    b[s[i].bj] = query(1, s[i].pos);
  }
  for (register int i = 1; i <= m; i++) printf("%lld\n", b[i]);
  return 0;
}

```