#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-all-loops")
using namespace std;
ifstream in;
ofstream out;
const long long kk = 1000;
const long long ml = kk * kk;
const long long mod = ml * kk + 7;
const long long inf = ml * ml * ml + 7;
const int s = 300;
int n, i, j;
vector<int> m;
vector<pair<int, int>> tos;
vector<vector<pair<int, int>>> allns;
vector<vector<int>> bs, all;
bool viv = false;
vector<int> merge(vector<int> a, vector<int> b) {
  int i = 0, j = 0;
  vector<int> res;
  while (i < a.size() && j < b.size()) {
    if (a[i] <= b[j])
      res.push_back(a[i]), i++;
    else
      res.push_back(b[j]), j++;
  }
  while (i < a.size()) res.push_back(a[i]), i++;
  while (j < b.size()) res.push_back(b[j]), j++;
  return res;
}
void prepare() {
  for (int i = 0; i < n; i++) {
    tos.push_back({m[i], -i});
  }
  sort(tos.rbegin(), tos.rend());
  for (auto &i : tos) i.second *= -1;
  for (int i = 0; i < n; i += s) {
    int r = i + s;
    r = min(r, n);
    vector<pair<int, int>> sv;
    for (int j = i; j < r; j++) sv.push_back(tos[j]);
    allns.push_back(sv);
    vector<int> vv;
    for (auto i : sv) vv.push_back(i.second);
    sort(vv.begin(), vv.end());
    all.push_back(vv);
  }
  for (int i = 0; i < all.size(); i++) {
    auto z = all[i];
    if (i) z = merge(bs[i - 1], z);
    bs.push_back(z);
    if (viv) {
      for (auto i : z) cout << i << ' ';
      cout << endl;
    }
  }
}
void solve() {
  int k, pos;
  cin >> k >> pos;
  pos--;
  int full = k / s;
  k -= full * s;
  vector<pair<int, int>> add2;
  for (int i = 0; i < k; i++) add2.push_back(allns[full][i]);
  vector<int> add;
  for (auto i : add2) add.push_back(i.second);
  sort(add.begin(), add.end());
  if (viv) {
    cout << "A ";
    for (auto i : add) cout << i << ' ';
    cout << endl;
  }
  if (full == 0) {
    if (viv) cout << "pos " << pos << ' ';
    cout << m[add[pos]] << '\n';
    return;
  }
  int pl = pos - add.size();
  if (pl < 0) {
    vector<int> a;
    for (int i = 0; i < min(bs[full - 1].size(), add.size()); i++)
      a.push_back(bs[full - 1][i]);
    a = merge(a, add);
    cout << m[a[pos]] << '\n';
    return;
  }
  if (viv) {
    cout << "B " << full << '\n';
    for (auto i : bs[full - 1]) cout << i << ' ';
    cout << '\n';
    for (auto i : add) cout << i << ' ';
    cout << '\n';
  }
  for (int i = 0; i < add.size(); i++) {
    if (bs[full - 1][pl] > add[i]) pl++;
  }
  cout << m[bs[full - 1][pl]] << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (i = 0; i < n; i++) {
    int a;
    cin >> a;
    m.push_back(a);
  }
  prepare();
  int q;
  cin >> q;
  while (q--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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("O3", "unroll-all-loops")
using namespace std;
ifstream in;
ofstream out;
const long long kk = 1000;
const long long ml = kk * kk;
const long long mod = ml * kk + 7;
const long long inf = ml * ml * ml + 7;
const int s = 300;
int n, i, j;
vector<int> m;
vector<pair<int, int>> tos;
vector<vector<pair<int, int>>> allns;
vector<vector<int>> bs, all;
bool viv = false;
vector<int> merge(vector<int> a, vector<int> b) {
  int i = 0, j = 0;
  vector<int> res;
  while (i < a.size() && j < b.size()) {
    if (a[i] <= b[j])
      res.push_back(a[i]), i++;
    else
      res.push_back(b[j]), j++;
  }
  while (i < a.size()) res.push_back(a[i]), i++;
  while (j < b.size()) res.push_back(b[j]), j++;
  return res;
}
void prepare() {
  for (int i = 0; i < n; i++) {
    tos.push_back({m[i], -i});
  }
  sort(tos.rbegin(), tos.rend());
  for (auto &i : tos) i.second *= -1;
  for (int i = 0; i < n; i += s) {
    int r = i + s;
    r = min(r, n);
    vector<pair<int, int>> sv;
    for (int j = i; j < r; j++) sv.push_back(tos[j]);
    allns.push_back(sv);
    vector<int> vv;
    for (auto i : sv) vv.push_back(i.second);
    sort(vv.begin(), vv.end());
    all.push_back(vv);
  }
  for (int i = 0; i < all.size(); i++) {
    auto z = all[i];
    if (i) z = merge(bs[i - 1], z);
    bs.push_back(z);
    if (viv) {
      for (auto i : z) cout << i << ' ';
      cout << endl;
    }
  }
}
void solve() {
  int k, pos;
  cin >> k >> pos;
  pos--;
  int full = k / s;
  k -= full * s;
  vector<pair<int, int>> add2;
  for (int i = 0; i < k; i++) add2.push_back(allns[full][i]);
  vector<int> add;
  for (auto i : add2) add.push_back(i.second);
  sort(add.begin(), add.end());
  if (viv) {
    cout << "A ";
    for (auto i : add) cout << i << ' ';
    cout << endl;
  }
  if (full == 0) {
    if (viv) cout << "pos " << pos << ' ';
    cout << m[add[pos]] << '\n';
    return;
  }
  int pl = pos - add.size();
  if (pl < 0) {
    vector<int> a;
    for (int i = 0; i < min(bs[full - 1].size(), add.size()); i++)
      a.push_back(bs[full - 1][i]);
    a = merge(a, add);
    cout << m[a[pos]] << '\n';
    return;
  }
  if (viv) {
    cout << "B " << full << '\n';
    for (auto i : bs[full - 1]) cout << i << ' ';
    cout << '\n';
    for (auto i : add) cout << i << ' ';
    cout << '\n';
  }
  for (int i = 0; i < add.size(); i++) {
    if (bs[full - 1][pl] > add[i]) pl++;
  }
  cout << m[bs[full - 1][pl]] << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (i = 0; i < n; i++) {
    int a;
    cin >> a;
    m.push_back(a);
  }
  prepare();
  int q;
  cin >> q;
  while (q--) solve();
  return 0;
}

```