#include <bits/stdc++.h>
using namespace std;
using namespace std;
template <typename T, typename U>
std::istream& operator>>(std::istream& i, pair<T, U>& p) {
  i >> p.first >> p.second;
  return i;
}
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T, typename U>
std::ostream& operator<<(std::ostream& o, const pair<T, U>& p) {
  o << p.first << ' ' << p.second;
  return o;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  if (t.empty()) o << '\n';
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T, vector<T>, less<T>>;
template <typename T>
bool in(T a, T b, T c) {
  return a <= b && b < c;
}
unsigned int logceil(int x) {
  return (unsigned int)(8 * sizeof(int) - __builtin_clz(x));
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.x) ^ (u(p.y) << 7);
  }
};
}  // namespace std
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <typename T>
class vector2 : public vector<vector<T>> {
 public:
  vector2() {}
  vector2(size_t a, size_t b, T t = T())
      : vector<vector<T>>(a, vector<T>(b, t)) {}
};
template <typename T>
class vector3 : public vector<vector2<T>> {
 public:
  vector3() {}
  vector3(size_t a, size_t b, size_t c, T t = T())
      : vector<vector2<T>>(a, vector2<T>(b, c, t)) {}
};
template <typename T>
class vector4 : public vector<vector3<T>> {
 public:
  vector4() {}
  vector4(size_t a, size_t b, size_t c, size_t d, T t = T())
      : vector<vector3<T>>(a, vector3<T>(b, c, d, t)) {}
};
template <typename T>
class vector5 : public vector<vector4<T>> {
 public:
  vector5() {}
  vector5(size_t a, size_t b, size_t c, size_t d, size_t e, T t = T())
      : vector<vector4<T>>(a, vector4<T>(b, c, d, e, t)) {}
};
void fastIO() { ios::sync_with_stdio(false); }
const int MAXN = 200200;
vector<vector<int>> t;
int n;
void add(int i, int v) {
  for (; i < n; i = (i | (i + 1))) t[i].push_back(v);
}
int cntd(int pos, int p) {
  return (lower_bound(t[pos].begin(), t[pos].end(), p) - t[pos].begin());
}
int cnt(int r, int p) {
  int result = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) result += cntd(r, p);
  return result;
}
class B1OptimalniePodposledovatelnostiUproshchyonnayaVersiya {
 public:
  void solve(std::istream& in, std::ostream& out) {
    fastIO();
    in >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) a[i] = rand();
    in >> a;
    vector<int> v;
    v.clear();
    for (int i = 0; i < n; ++i) {
      v.push_back(i);
    }
    sort(v.begin(), v.end(), [&a](int x, int y) {
      return pair<int, int>(a[x], -x) > pair<int, int>(a[y], -y);
    });
    t.clear();
    t.resize(n);
    for (int i = 0; i < n; ++i) add(i, v[i]);
    for (int i = 0; i < n; ++i) {
      sort(t[i].begin(), t[i].end());
    }
    int m, k, pos;
    in >> m;
    for (int _ = 0; _ < m; ++_) {
      k = n - 1;
      pos = rand() % n + 1;
      in >> k >> pos;
      int l = 0;
      int r = n + 1;
      while (r - l > 1) {
        int md = (r + l) / 2;
        if (cnt(k - 1, md) >= pos)
          r = md;
        else
          l = md;
      }
      out << a[r - 1] << endl;
    }
  }
};
int main() {
  B1OptimalniePodposledovatelnostiUproshchyonnayaVersiya solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  cerr << 1.0 * clock() / CLOCKS_PER_SEC << endl;
  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;
using namespace std;
template <typename T, typename U>
std::istream& operator>>(std::istream& i, pair<T, U>& p) {
  i >> p.first >> p.second;
  return i;
}
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T, typename U>
std::ostream& operator<<(std::ostream& o, const pair<T, U>& p) {
  o << p.first << ' ' << p.second;
  return o;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  if (t.empty()) o << '\n';
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T, vector<T>, less<T>>;
template <typename T>
bool in(T a, T b, T c) {
  return a <= b && b < c;
}
unsigned int logceil(int x) {
  return (unsigned int)(8 * sizeof(int) - __builtin_clz(x));
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.x) ^ (u(p.y) << 7);
  }
};
}  // namespace std
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <typename T>
class vector2 : public vector<vector<T>> {
 public:
  vector2() {}
  vector2(size_t a, size_t b, T t = T())
      : vector<vector<T>>(a, vector<T>(b, t)) {}
};
template <typename T>
class vector3 : public vector<vector2<T>> {
 public:
  vector3() {}
  vector3(size_t a, size_t b, size_t c, T t = T())
      : vector<vector2<T>>(a, vector2<T>(b, c, t)) {}
};
template <typename T>
class vector4 : public vector<vector3<T>> {
 public:
  vector4() {}
  vector4(size_t a, size_t b, size_t c, size_t d, T t = T())
      : vector<vector3<T>>(a, vector3<T>(b, c, d, t)) {}
};
template <typename T>
class vector5 : public vector<vector4<T>> {
 public:
  vector5() {}
  vector5(size_t a, size_t b, size_t c, size_t d, size_t e, T t = T())
      : vector<vector4<T>>(a, vector4<T>(b, c, d, e, t)) {}
};
void fastIO() { ios::sync_with_stdio(false); }
const int MAXN = 200200;
vector<vector<int>> t;
int n;
void add(int i, int v) {
  for (; i < n; i = (i | (i + 1))) t[i].push_back(v);
}
int cntd(int pos, int p) {
  return (lower_bound(t[pos].begin(), t[pos].end(), p) - t[pos].begin());
}
int cnt(int r, int p) {
  int result = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) result += cntd(r, p);
  return result;
}
class B1OptimalniePodposledovatelnostiUproshchyonnayaVersiya {
 public:
  void solve(std::istream& in, std::ostream& out) {
    fastIO();
    in >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) a[i] = rand();
    in >> a;
    vector<int> v;
    v.clear();
    for (int i = 0; i < n; ++i) {
      v.push_back(i);
    }
    sort(v.begin(), v.end(), [&a](int x, int y) {
      return pair<int, int>(a[x], -x) > pair<int, int>(a[y], -y);
    });
    t.clear();
    t.resize(n);
    for (int i = 0; i < n; ++i) add(i, v[i]);
    for (int i = 0; i < n; ++i) {
      sort(t[i].begin(), t[i].end());
    }
    int m, k, pos;
    in >> m;
    for (int _ = 0; _ < m; ++_) {
      k = n - 1;
      pos = rand() % n + 1;
      in >> k >> pos;
      int l = 0;
      int r = n + 1;
      while (r - l > 1) {
        int md = (r + l) / 2;
        if (cnt(k - 1, md) >= pos)
          r = md;
        else
          l = md;
      }
      out << a[r - 1] << endl;
    }
  }
};
int main() {
  B1OptimalniePodposledovatelnostiUproshchyonnayaVersiya solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  cerr << 1.0 * clock() / CLOCKS_PER_SEC << endl;
  return 0;
}

```