#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using mii = map<int, int>;
const double Eps = 1e-8;
const int Inf = 1e9 + 9;
const int Mod = 1e9 + 7;
const int N = 1e5 + 9;
const int dx4[4] = {0, 1, 0, -1};
const int dy4[4] = {-1, 0, 1, 0};
const int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
constexpr double pi = 3.141592653589793238462643383279502884L;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
void prime_siever(long long p) {
  bool prime[p + 9];
  for (long long i = 2; i * i <= p; i++)
    if (prime[i] == 0)
      for (long long j = i * i; j <= p; j += i) prime[j] = 1;
}
int mul(int a, int b) { return (1LL * a * b) % Mod; }
int add(int a, int b) {
  a += b;
  if (a >= Mod) a -= Mod;
  if (a < 0) a += Mod;
  return a;
}
long long bin_power(long long a, long long n) {
  if (a == 0) return 0;
  long long res = 1;
  while (n) {
    if (n % 2) {
      res = (res * a) % Mod;
      n--;
    } else {
      a = (a * a) % Mod;
      n /= 2;
    }
  }
  return res;
}
vector<int> a;
vector<pair<int, int> > b, c;
priority_queue<int, vector<int>, greater<int> > qi;
map<pair<int, int>, int> mp;
void solve() {
  int q, n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a.push_back(x);
    qi.push(a[i]);
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    b.push_back({x, y});
  }
  c = b;
  sort(b.rbegin(), b.rend());
  for (int i = 0; i < q; i++) {
    int siz = b[i].first, idx = b[i].second - 1;
    while (qi.size() != siz) {
      reverse(a.begin(), a.end());
      a.erase(find(a.begin(), a.end(), qi.top()));
      reverse(a.begin(), a.end());
      qi.pop();
    }
    mp[{siz, idx + 1}] = a[idx];
  }
  for (int i = 0; i < q; i++) {
    cout << mp[c[i]] << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
using vi = vector<int>;
using mii = map<int, int>;
const double Eps = 1e-8;
const int Inf = 1e9 + 9;
const int Mod = 1e9 + 7;
const int N = 1e5 + 9;
const int dx4[4] = {0, 1, 0, -1};
const int dy4[4] = {-1, 0, 1, 0};
const int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
constexpr double pi = 3.141592653589793238462643383279502884L;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
void prime_siever(long long p) {
  bool prime[p + 9];
  for (long long i = 2; i * i <= p; i++)
    if (prime[i] == 0)
      for (long long j = i * i; j <= p; j += i) prime[j] = 1;
}
int mul(int a, int b) { return (1LL * a * b) % Mod; }
int add(int a, int b) {
  a += b;
  if (a >= Mod) a -= Mod;
  if (a < 0) a += Mod;
  return a;
}
long long bin_power(long long a, long long n) {
  if (a == 0) return 0;
  long long res = 1;
  while (n) {
    if (n % 2) {
      res = (res * a) % Mod;
      n--;
    } else {
      a = (a * a) % Mod;
      n /= 2;
    }
  }
  return res;
}
vector<int> a;
vector<pair<int, int> > b, c;
priority_queue<int, vector<int>, greater<int> > qi;
map<pair<int, int>, int> mp;
void solve() {
  int q, n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a.push_back(x);
    qi.push(a[i]);
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    b.push_back({x, y});
  }
  c = b;
  sort(b.rbegin(), b.rend());
  for (int i = 0; i < q; i++) {
    int siz = b[i].first, idx = b[i].second - 1;
    while (qi.size() != siz) {
      reverse(a.begin(), a.end());
      a.erase(find(a.begin(), a.end(), qi.top()));
      reverse(a.begin(), a.end());
      qi.pop();
    }
    mp[{siz, idx + 1}] = a[idx];
  }
  for (int i = 0; i < q; i++) {
    cout << mp[c[i]] << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

```