#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
using namespace std;
const long long INF = 1e18;
const int int_INF = 1e9 + 13;
const long double PI = acos(-1);
const long double EPS = 1e-9;
const long long MOD = 1e9 + 7;
const long long MAXN = 2e5 + 13;
vector<int> tree(4 * MAXN);
vector<int> arr(MAXN);
void build(int cur, int l, int r) {
  if (l == r - 1)
    tree[cur] = 0;
  else if (l >= r)
    return;
  else {
    long long m = (l + r) / 2;
    build(2 * cur + 1, l, m);
    build(2 * cur + 2, m, r);
    tree[cur] = tree[2 * cur + 1] + tree[2 * cur + 2];
  }
}
void add(int cur, int l, int r, int x, int pos) {
  if (l >= r || l > pos || pos >= r)
    return;
  else if (l == r - 1)
    tree[cur] += x;
  else {
    long long m = (l + r) / 2;
    add(2 * cur + 1, l, m, x, pos);
    add(2 * cur + 2, m, r, x, pos);
    tree[cur] += x;
  }
}
int with_sum(int cur, int l, int r, int k) {
  if (l == r - 1)
    return l;
  else if (tree[2 * cur + 1] >= k)
    return with_sum(2 * cur + 1, l, (l + r) / 2, k);
  else
    return with_sum(2 * cur + 2, (l + r) / 2, r, k - tree[2 * cur + 1]);
}
void solve() {
  int n, m;
  cin >> n;
  vector<int> a(n);
  for (int &x : a) cin >> x;
  for (int &x : tree) x = 0;
  vector<pair<int, int> > precalc;
  vector<tuple<int, int, int> > queries;
  for (int i = 0; i < n; ++i) precalc.emplace_back(a[i], -i);
  sort(precalc.begin(), precalc.end());
  reverse(precalc.begin(), precalc.end());
  for (int i = 0; i < n; ++i) arr[i] = abs(precalc[i].second) + 1;
  cin >> m;
  vector<int> res(m);
  for (int q = 0; q < m; ++q) {
    int k, j;
    cin >> k >> j;
    queries.emplace_back(k, j, q);
  }
  sort(queries.begin(), queries.end());
  build(0, 0, n + 1);
  long long cur_pos = -1;
  for (int t = 0; t < m; ++t) {
    auto [k, j, q] = queries[t];
    for (int i = cur_pos + 1; i <= k - 1; ++i) add(0, 0, n + 1, 1, arr[i]);
    res[q] = a[with_sum(0, 0, n + 1, j) - 1];
    cur_pos = k - 1;
  }
  for (int x : res) cout << x << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  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>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
using namespace std;
const long long INF = 1e18;
const int int_INF = 1e9 + 13;
const long double PI = acos(-1);
const long double EPS = 1e-9;
const long long MOD = 1e9 + 7;
const long long MAXN = 2e5 + 13;
vector<int> tree(4 * MAXN);
vector<int> arr(MAXN);
void build(int cur, int l, int r) {
  if (l == r - 1)
    tree[cur] = 0;
  else if (l >= r)
    return;
  else {
    long long m = (l + r) / 2;
    build(2 * cur + 1, l, m);
    build(2 * cur + 2, m, r);
    tree[cur] = tree[2 * cur + 1] + tree[2 * cur + 2];
  }
}
void add(int cur, int l, int r, int x, int pos) {
  if (l >= r || l > pos || pos >= r)
    return;
  else if (l == r - 1)
    tree[cur] += x;
  else {
    long long m = (l + r) / 2;
    add(2 * cur + 1, l, m, x, pos);
    add(2 * cur + 2, m, r, x, pos);
    tree[cur] += x;
  }
}
int with_sum(int cur, int l, int r, int k) {
  if (l == r - 1)
    return l;
  else if (tree[2 * cur + 1] >= k)
    return with_sum(2 * cur + 1, l, (l + r) / 2, k);
  else
    return with_sum(2 * cur + 2, (l + r) / 2, r, k - tree[2 * cur + 1]);
}
void solve() {
  int n, m;
  cin >> n;
  vector<int> a(n);
  for (int &x : a) cin >> x;
  for (int &x : tree) x = 0;
  vector<pair<int, int> > precalc;
  vector<tuple<int, int, int> > queries;
  for (int i = 0; i < n; ++i) precalc.emplace_back(a[i], -i);
  sort(precalc.begin(), precalc.end());
  reverse(precalc.begin(), precalc.end());
  for (int i = 0; i < n; ++i) arr[i] = abs(precalc[i].second) + 1;
  cin >> m;
  vector<int> res(m);
  for (int q = 0; q < m; ++q) {
    int k, j;
    cin >> k >> j;
    queries.emplace_back(k, j, q);
  }
  sort(queries.begin(), queries.end());
  build(0, 0, n + 1);
  long long cur_pos = -1;
  for (int t = 0; t < m; ++t) {
    auto [k, j, q] = queries[t];
    for (int i = cur_pos + 1; i <= k - 1; ++i) add(0, 0, n + 1, 1, arr[i]);
    res[q] = a[with_sum(0, 0, n + 1, j) - 1];
    cur_pos = k - 1;
  }
  for (int x : res) cout << x << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

```