#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
using ll = long long;
pair<ll, int> a[maxn];
ll b[maxn];
struct NODE {
  NODE *left, *right;
  int val;
  NODE() {
    left = nullptr;
    right = nullptr;
    val = 0;
  }
} * root[maxn];
void build(NODE *node, int b, int e) {
  if (b == e) {
    node->val = 0;
    return;
  }
  int mid = (b + e) >> 1;
  if (node->left == nullptr) node->left = new NODE();
  if (node->right == nullptr) node->right = new NODE();
  build(node->left, b, mid);
  build(node->right, mid + 1, e);
  node->val = node->left->val + node->right->val;
}
void upd(NODE *prv, NODE *cur, int b, int e, int idx) {
  if (idx > e || idx < b) return;
  if (idx == b && idx == e) {
    cur->val = prv->val + 1;
    return;
  }
  int mid = (b + e) >> 1;
  if (idx <= mid) {
    if (cur->left == nullptr) cur->left = new NODE();
    cur->right = prv->right;
    upd(prv->left, cur->left, b, mid, idx);
  } else {
    if (cur->right == nullptr) cur->right = new NODE();
    cur->left = prv->left;
    upd(prv->right, cur->right, mid + 1, e, idx);
  }
  cur->val = cur->left->val + cur->right->val;
}
int query(NODE *prv, NODE *cur, int b, int e, int k) {
  if (b == e) return b;
  int mid = (b + e) >> 1;
  if (k <= cur->left->val - prv->left->val)
    return query(prv->left, cur->left, b, mid, k);
  else {
    return query(prv->right, cur->right, mid + 1, e,
                 k - (cur->left->val - prv->left->val));
  }
}
int main(int argc, char const *argv[]) {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i].first);
    b[i] = a[i].first;
    a[i].second = i;
  }
  sort(a + 1, a + n + 1, [&](pair<ll, int> x, pair<ll, int> y) {
    if (x.first != y.first)
      return x.first > y.first;
    else
      return x.second < y.second;
  });
  root[0] = new NODE();
  build(root[0], 1, n);
  for (int i = 1; i <= n; ++i) {
    root[i] = new NODE();
    upd(root[i - 1], root[i], 1, n, a[i].second);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int k, pos;
    scanf("%d %d", &k, &pos);
    printf("%lld\n", b[query(root[0], root[k], 1, n, pos)]);
  }
  return 0;
}

### Prompt

Your task is to create 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 int maxn = 200005;
using ll = long long;
pair<ll, int> a[maxn];
ll b[maxn];
struct NODE {
  NODE *left, *right;
  int val;
  NODE() {
    left = nullptr;
    right = nullptr;
    val = 0;
  }
} * root[maxn];
void build(NODE *node, int b, int e) {
  if (b == e) {
    node->val = 0;
    return;
  }
  int mid = (b + e) >> 1;
  if (node->left == nullptr) node->left = new NODE();
  if (node->right == nullptr) node->right = new NODE();
  build(node->left, b, mid);
  build(node->right, mid + 1, e);
  node->val = node->left->val + node->right->val;
}
void upd(NODE *prv, NODE *cur, int b, int e, int idx) {
  if (idx > e || idx < b) return;
  if (idx == b && idx == e) {
    cur->val = prv->val + 1;
    return;
  }
  int mid = (b + e) >> 1;
  if (idx <= mid) {
    if (cur->left == nullptr) cur->left = new NODE();
    cur->right = prv->right;
    upd(prv->left, cur->left, b, mid, idx);
  } else {
    if (cur->right == nullptr) cur->right = new NODE();
    cur->left = prv->left;
    upd(prv->right, cur->right, mid + 1, e, idx);
  }
  cur->val = cur->left->val + cur->right->val;
}
int query(NODE *prv, NODE *cur, int b, int e, int k) {
  if (b == e) return b;
  int mid = (b + e) >> 1;
  if (k <= cur->left->val - prv->left->val)
    return query(prv->left, cur->left, b, mid, k);
  else {
    return query(prv->right, cur->right, mid + 1, e,
                 k - (cur->left->val - prv->left->val));
  }
}
int main(int argc, char const *argv[]) {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i].first);
    b[i] = a[i].first;
    a[i].second = i;
  }
  sort(a + 1, a + n + 1, [&](pair<ll, int> x, pair<ll, int> y) {
    if (x.first != y.first)
      return x.first > y.first;
    else
      return x.second < y.second;
  });
  root[0] = new NODE();
  build(root[0], 1, n);
  for (int i = 1; i <= n; ++i) {
    root[i] = new NODE();
    upd(root[i - 1], root[i], 1, n, a[i].second);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int k, pos;
    scanf("%d %d", &k, &pos);
    printf("%lld\n", b[query(root[0], root[k], 1, n, pos)]);
  }
  return 0;
}

```