#include <bits/stdc++.h>
using namespace std;
namespace kotespace {
template <class T>
class duplet {
 private:
 public:
  T x, y;
  duplet(){};
  duplet(T a, T b) : x(a), y(b){};
  bool operator<(const duplet P) const {
    return (x < P.x || (x == P.x && y < P.y));
  }
  bool operator>(const duplet P) const {
    return (x > P.x || (x == P.x && y > P.y));
  }
  bool operator==(const duplet P) const { return (x == P.x && y == P.y); }
  bool operator!=(const duplet P) const { return (x != P.x || y != P.y); }
  void reverse() { std::swap(x, y); }
};
template <class P>
istream &operator>>(istream &in, duplet<P> &T) {
  return (in >> T.x >> T.y);
}
template <class P>
ostream &operator<<(ostream &out, duplet<P> T) {
  return (out << T.x << " " << T.y);
}
template <class T>
class triplet {
 private:
 public:
  T x, y, z;
  triplet(){};
  triplet(T a, T b, T c) : x(a), y(b), z(c){};
  bool operator<(const triplet P) const {
    return (x < P.x || (x == P.x && y < P.y) ||
            (x == P.x && y == P.y && z < P.z));
  }
  bool operator>(const triplet P) const {
    return (x > P.x || (x == P.x && y > P.y) ||
            (x == P.x && y == P.y && z > P.z));
  }
  bool operator==(const triplet P) const {
    return (x == P.x && y == P.y && z == P.z);
  }
  bool operator!=(const triplet P) const {
    return (x != P.x || y != P.y || z != P.z);
  }
  void reverse() { std::swap(x, z); }
  void cycle_right(int a) {
    if (a == 1) {
      std::swap(x, y);
      std::swap(x, z);
    }
    if (a == 2) {
      std::swap(x, z);
      std::swap(y, x);
    }
  }
  void cycle_left(int a) {
    if (a == 1) {
      std::swap(x, z);
      std::swap(y, x);
    }
    if (a == 2) {
      std::swap(x, y);
      std::swap(x, z);
    }
  }
};
template <class P>
istream &operator>>(istream &in, triplet<P> &T) {
  return (in >> T.x >> T.y >> T.z);
}
template <class P>
ostream &operator<<(ostream &out, triplet<P> &T) {
  return (out << T.x << " " << T.y << " " << T.z);
}
}  // namespace kotespace
using namespace kotespace;
long long inf = 1000 * 1000 * 1000 + 5;
long long inf64 = inf * inf;
long long mod = 228228227;
vector<duplet<int> > a;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cerr << fixed << setprecision(10);
  cout << fixed << setprecision(10);
  srand(time(0));
  float START_TIME = clock();
  int n;
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].x;
    a[i].x *= -1;
    a[i].y = i;
  }
  sort(a.begin(), a.end());
  int k;
  cin >> k;
  for (int i = 0; i < k; ++i) {
    int x, y;
    cin >> x >> y;
    vector<duplet<int> > b;
    for (int j = 0; j < x; ++j) {
      b.emplace_back(a[j].y, -a[j].x);
    }
    sort(b.begin(), b.end());
    cout << b[y - 1].y << endl;
  }
  cerr << endl << (clock() - START_TIME) / CLOCKS_PER_SEC << " sec." << endl;
}

### Prompt

Generate 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;
namespace kotespace {
template <class T>
class duplet {
 private:
 public:
  T x, y;
  duplet(){};
  duplet(T a, T b) : x(a), y(b){};
  bool operator<(const duplet P) const {
    return (x < P.x || (x == P.x && y < P.y));
  }
  bool operator>(const duplet P) const {
    return (x > P.x || (x == P.x && y > P.y));
  }
  bool operator==(const duplet P) const { return (x == P.x && y == P.y); }
  bool operator!=(const duplet P) const { return (x != P.x || y != P.y); }
  void reverse() { std::swap(x, y); }
};
template <class P>
istream &operator>>(istream &in, duplet<P> &T) {
  return (in >> T.x >> T.y);
}
template <class P>
ostream &operator<<(ostream &out, duplet<P> T) {
  return (out << T.x << " " << T.y);
}
template <class T>
class triplet {
 private:
 public:
  T x, y, z;
  triplet(){};
  triplet(T a, T b, T c) : x(a), y(b), z(c){};
  bool operator<(const triplet P) const {
    return (x < P.x || (x == P.x && y < P.y) ||
            (x == P.x && y == P.y && z < P.z));
  }
  bool operator>(const triplet P) const {
    return (x > P.x || (x == P.x && y > P.y) ||
            (x == P.x && y == P.y && z > P.z));
  }
  bool operator==(const triplet P) const {
    return (x == P.x && y == P.y && z == P.z);
  }
  bool operator!=(const triplet P) const {
    return (x != P.x || y != P.y || z != P.z);
  }
  void reverse() { std::swap(x, z); }
  void cycle_right(int a) {
    if (a == 1) {
      std::swap(x, y);
      std::swap(x, z);
    }
    if (a == 2) {
      std::swap(x, z);
      std::swap(y, x);
    }
  }
  void cycle_left(int a) {
    if (a == 1) {
      std::swap(x, z);
      std::swap(y, x);
    }
    if (a == 2) {
      std::swap(x, y);
      std::swap(x, z);
    }
  }
};
template <class P>
istream &operator>>(istream &in, triplet<P> &T) {
  return (in >> T.x >> T.y >> T.z);
}
template <class P>
ostream &operator<<(ostream &out, triplet<P> &T) {
  return (out << T.x << " " << T.y << " " << T.z);
}
}  // namespace kotespace
using namespace kotespace;
long long inf = 1000 * 1000 * 1000 + 5;
long long inf64 = inf * inf;
long long mod = 228228227;
vector<duplet<int> > a;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cerr << fixed << setprecision(10);
  cout << fixed << setprecision(10);
  srand(time(0));
  float START_TIME = clock();
  int n;
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].x;
    a[i].x *= -1;
    a[i].y = i;
  }
  sort(a.begin(), a.end());
  int k;
  cin >> k;
  for (int i = 0; i < k; ++i) {
    int x, y;
    cin >> x >> y;
    vector<duplet<int> > b;
    for (int j = 0; j < x; ++j) {
      b.emplace_back(a[j].y, -a[j].x);
    }
    sort(b.begin(), b.end());
    cout << b[y - 1].y << endl;
  }
  cerr << endl << (clock() - START_TIME) / CLOCKS_PER_SEC << " sec." << endl;
}

```