#include <bits/stdc++.h>
using namespace std;
const long long int inf = 1e18;
const long long int mod = 1e9 + 7;
const long long int MOD = 998244353;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug &operator<<(const c &) {
    return *this;
  }
};
vector<char *> tokenizer(const char *args) {
  char *token = new char[111];
  strcpy(token, args);
  token = strtok(token, ", ");
  vector<char *> v({token});
  while ((token = strtok(NULL, ", "))) v.push_back(token);
  return reverse(v.begin(), v.end()), v;
}
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wunused-parameter"
void debugg(vector<char *> args) { cerr << "\b\b "; }
#pragma clang diagnostic pop
template <typename Head, typename... Tail>
void debugg(vector<char *> args, Head H, Tail... T) {
  debug() << " [" << args.back() << ": " << H << "] ";
  args.pop_back();
  debugg(args, T...);
}
template <typename T>
T power(T a, T b) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    T res = (power(a, b / 2));
    if (b % 2) {
      return (res * res * a);
    } else {
      return res * res;
    }
  }
}
template <typename T>
T power(T a, T b, T modulo) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    T res = (power(a, b / 2, modulo) % modulo);
    if (b % 2) {
      return ((((res % modulo) * (res % modulo)) % modulo) * (a % modulo)) %
             modulo;
    } else {
      return ((res % modulo) * (res % modulo)) % modulo;
    }
  }
}
template <typename T>
T gcd(T a, T b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
void factorial(vector<long long int> &fact, long long n) {
  fact.resize(n + 1, 1);
  fact[0] = 1;
  fact[1] = 1;
  for (int i = 2; i <= n; ++i) {
    fact[i] = ((fact[i - 1] % mod) * (i % mod)) % mod;
  }
}
long long mod_inv(long long a) {
  return (power<long long>(a, mod - 2, mod)) % mod;
}
long long ncr(long long n, long long r, vector<long long int> &fact) {
  if (r > n or n < 0 or r < 0) return 0LL;
  return (((fact[n] % mod) * (mod_inv(fact[n - r]) % mod)) % mod *
          (mod_inv(fact[r]) % mod)) %
         mod;
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n;
  cin >> n;
  vector<long long int> v(n - 1 + 1);
  for (auto i = 0; i <= n - 1; i++) {
    cin >> v[i];
  };
  {};
  vector<pair<long long, long long>> s;
  for (int i = 0; i < n; ++i) {
    s.push_back(pair<long long, long long>{v[i], i + 1});
  }
  sort((s).begin(), (s).end(), cmp);
  {};
  int t;
  cin >> t;
  while (t--) {
    long long k, p;
    cin >> k >> p;
    vector<long long int> a;
    for (int i = 0; i < k; ++i) {
      a.push_back(s[i].second);
    }
    sort((a).begin(), (a).end());
    {};
    cout << ((v[a[p - 1] - 1])) << "\n";
    ;
  }
  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 long long int inf = 1e18;
const long long int mod = 1e9 + 7;
const long long int MOD = 998244353;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug &operator<<(const c &) {
    return *this;
  }
};
vector<char *> tokenizer(const char *args) {
  char *token = new char[111];
  strcpy(token, args);
  token = strtok(token, ", ");
  vector<char *> v({token});
  while ((token = strtok(NULL, ", "))) v.push_back(token);
  return reverse(v.begin(), v.end()), v;
}
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wunused-parameter"
void debugg(vector<char *> args) { cerr << "\b\b "; }
#pragma clang diagnostic pop
template <typename Head, typename... Tail>
void debugg(vector<char *> args, Head H, Tail... T) {
  debug() << " [" << args.back() << ": " << H << "] ";
  args.pop_back();
  debugg(args, T...);
}
template <typename T>
T power(T a, T b) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    T res = (power(a, b / 2));
    if (b % 2) {
      return (res * res * a);
    } else {
      return res * res;
    }
  }
}
template <typename T>
T power(T a, T b, T modulo) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    T res = (power(a, b / 2, modulo) % modulo);
    if (b % 2) {
      return ((((res % modulo) * (res % modulo)) % modulo) * (a % modulo)) %
             modulo;
    } else {
      return ((res % modulo) * (res % modulo)) % modulo;
    }
  }
}
template <typename T>
T gcd(T a, T b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
void factorial(vector<long long int> &fact, long long n) {
  fact.resize(n + 1, 1);
  fact[0] = 1;
  fact[1] = 1;
  for (int i = 2; i <= n; ++i) {
    fact[i] = ((fact[i - 1] % mod) * (i % mod)) % mod;
  }
}
long long mod_inv(long long a) {
  return (power<long long>(a, mod - 2, mod)) % mod;
}
long long ncr(long long n, long long r, vector<long long int> &fact) {
  if (r > n or n < 0 or r < 0) return 0LL;
  return (((fact[n] % mod) * (mod_inv(fact[n - r]) % mod)) % mod *
          (mod_inv(fact[r]) % mod)) %
         mod;
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n;
  cin >> n;
  vector<long long int> v(n - 1 + 1);
  for (auto i = 0; i <= n - 1; i++) {
    cin >> v[i];
  };
  {};
  vector<pair<long long, long long>> s;
  for (int i = 0; i < n; ++i) {
    s.push_back(pair<long long, long long>{v[i], i + 1});
  }
  sort((s).begin(), (s).end(), cmp);
  {};
  int t;
  cin >> t;
  while (t--) {
    long long k, p;
    cin >> k >> p;
    vector<long long int> a;
    for (int i = 0; i < k; ++i) {
      a.push_back(s[i].second);
    }
    sort((a).begin(), (a).end());
    {};
    cout << ((v[a[p - 1] - 1])) << "\n";
    ;
  }
  return 0;
}

```