#include <bits/stdc++.h>
using namespace std;
using Int = long long;
using ii = pair<Int, Int>;
template <class T>
inline bool chmax(T& a, const T& b) {
  if (a >= b) return false;
  a = b;
  return true;
}
template <class T>
inline bool chmin(T& a, const T& b) {
  if (a <= b) return false;
  a = b;
  return true;
}
struct SetupIO {
  SetupIO() {
    cin.tie(nullptr), ios::sync_with_stdio(false),
        cout << fixed << setprecision(20);
  }
} setup_io;
template <template <class, class...> class TT, class... Args>
ostream& operator<<(ostream& out, TT<Args...> const& cont) {
  for (auto&& elem : cont) out << elem << " ";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<vector<T>>& vs) {
  for (auto&& v : vs) cout << v << endl;
  return out;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> const& val) {
  return (out << "{" << val.first << " " << val.second << "}");
}
void dump_func() { cerr << endl; }
template <class Head, class... Tail>
void dump_func(Head&& h, Tail&&... t) {
  cerr << h << (sizeof...(Tail) == 0 ? "" : ", "),
      dump_func(forward<Tail>(t)...);
}
class BinaryIndexedTree {
  int max_pow;

 public:
  vector<int> bit;
  int N;
  BinaryIndexedTree(int n, int x = 0) : N(n), bit(n, x) {
    for (max_pow = 1; max_pow <= N / 2; max_pow *= 2)
      ;
  }
  BinaryIndexedTree(const vector<int>& a) : N(a.size()), bit(a) {
    for (max_pow = 1; max_pow <= N / 2; max_pow *= 2)
      ;
    for (Int i = 0, max_i = (N - 1); i < max_i; i++)
      if ((i | (i + 1)) < N) {
        bit[i | (i + 1)] += bit[i];
      }
  }
  void add(int k, int x) {
    for (int i = k; i < N; i |= i + 1) {
      bit[i] += x;
    }
  }
  int sum(int k) {
    int ret = 0;
    for (int i = k - 1; i >= 0; i = (i & (i + 1)) - 1) {
      ret += bit[i];
    }
    return ret;
  }
  int sum() { return sum(N); }
  int query(int a, int b) {
    assert(a <= b);
    return sum(b) - sum(a);
  }
  int operator[](int k) { return query(k, k + 1); }
  void update(int k, int x) { add(k, x - operator[](k)); }
  int lower_bound(int x) {
    int i = -1;
    for (int w = max_pow; w > 0; w /= 2) {
      if (i + w < N && bit[i + w] < x) {
        x -= bit[i + w];
        i += w;
      }
    }
    return i + 1;
  }
  int lower_bound(int begin, int x) {
    int lb = lower_bound(x + sum(begin));
    return max(begin, lb);
  }
  int upper_bound(int x) {
    int i = -1;
    for (int w = max_pow; w > 0; w /= 2) {
      if (i + w < N && bit[i + w] <= x) {
        x -= bit[i + w];
        i += w;
      }
    }
    return i + 1;
  }
  int upper_bound(int begin, int x) {
    int ub = upper_bound(x + sum(begin));
    return max(begin, ub);
  }
};
signed main() {
  Int n;
  cin >> n;
  vector<Int> a(n);
  for (Int i = 0, max_i = (n); i < max_i; i++) cin >> a[i];
  vector<Int> ord(n);
  for (Int i = 0, max_i = (n); i < max_i; i++) ord[i] = i;
  sort(begin(ord), end(ord),
       [&a](Int i, Int j) { return a[i] == a[j] ? i < j : a[i] > a[j]; });
  Int m;
  cin >> m;
  vector<Int> ks(m), pos(m);
  for (Int i = 0, max_i = (m); i < max_i; i++) cin >> ks[i] >> pos[i];
  vector<Int> qs(m);
  for (Int i = 0, max_i = (m); i < max_i; i++) qs[i] = i;
  sort(begin(qs), end(qs), [&](Int i, Int j) { return ks[i] < ks[j]; });
  vector<Int> ans(m);
  BinaryIndexedTree bit(n);
  Int curk = 0;
  for (Int q = 0, max_i = (m); q < max_i; q++) {
    Int k = ks[qs[q]], p = pos[qs[q]];
    while (curk < k) {
      assert(bit[ord[curk]] == 0);
      bit.add(ord[curk], 1);
      curk++;
    }
    Int idx = bit.lower_bound(p);
    ans[qs[q]] = a[idx];
  }
  for (Int x : ans) cout << x << '\n';
}

### Prompt

Please provide a Cpp coded solution 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 Int = long long;
using ii = pair<Int, Int>;
template <class T>
inline bool chmax(T& a, const T& b) {
  if (a >= b) return false;
  a = b;
  return true;
}
template <class T>
inline bool chmin(T& a, const T& b) {
  if (a <= b) return false;
  a = b;
  return true;
}
struct SetupIO {
  SetupIO() {
    cin.tie(nullptr), ios::sync_with_stdio(false),
        cout << fixed << setprecision(20);
  }
} setup_io;
template <template <class, class...> class TT, class... Args>
ostream& operator<<(ostream& out, TT<Args...> const& cont) {
  for (auto&& elem : cont) out << elem << " ";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<vector<T>>& vs) {
  for (auto&& v : vs) cout << v << endl;
  return out;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> const& val) {
  return (out << "{" << val.first << " " << val.second << "}");
}
void dump_func() { cerr << endl; }
template <class Head, class... Tail>
void dump_func(Head&& h, Tail&&... t) {
  cerr << h << (sizeof...(Tail) == 0 ? "" : ", "),
      dump_func(forward<Tail>(t)...);
}
class BinaryIndexedTree {
  int max_pow;

 public:
  vector<int> bit;
  int N;
  BinaryIndexedTree(int n, int x = 0) : N(n), bit(n, x) {
    for (max_pow = 1; max_pow <= N / 2; max_pow *= 2)
      ;
  }
  BinaryIndexedTree(const vector<int>& a) : N(a.size()), bit(a) {
    for (max_pow = 1; max_pow <= N / 2; max_pow *= 2)
      ;
    for (Int i = 0, max_i = (N - 1); i < max_i; i++)
      if ((i | (i + 1)) < N) {
        bit[i | (i + 1)] += bit[i];
      }
  }
  void add(int k, int x) {
    for (int i = k; i < N; i |= i + 1) {
      bit[i] += x;
    }
  }
  int sum(int k) {
    int ret = 0;
    for (int i = k - 1; i >= 0; i = (i & (i + 1)) - 1) {
      ret += bit[i];
    }
    return ret;
  }
  int sum() { return sum(N); }
  int query(int a, int b) {
    assert(a <= b);
    return sum(b) - sum(a);
  }
  int operator[](int k) { return query(k, k + 1); }
  void update(int k, int x) { add(k, x - operator[](k)); }
  int lower_bound(int x) {
    int i = -1;
    for (int w = max_pow; w > 0; w /= 2) {
      if (i + w < N && bit[i + w] < x) {
        x -= bit[i + w];
        i += w;
      }
    }
    return i + 1;
  }
  int lower_bound(int begin, int x) {
    int lb = lower_bound(x + sum(begin));
    return max(begin, lb);
  }
  int upper_bound(int x) {
    int i = -1;
    for (int w = max_pow; w > 0; w /= 2) {
      if (i + w < N && bit[i + w] <= x) {
        x -= bit[i + w];
        i += w;
      }
    }
    return i + 1;
  }
  int upper_bound(int begin, int x) {
    int ub = upper_bound(x + sum(begin));
    return max(begin, ub);
  }
};
signed main() {
  Int n;
  cin >> n;
  vector<Int> a(n);
  for (Int i = 0, max_i = (n); i < max_i; i++) cin >> a[i];
  vector<Int> ord(n);
  for (Int i = 0, max_i = (n); i < max_i; i++) ord[i] = i;
  sort(begin(ord), end(ord),
       [&a](Int i, Int j) { return a[i] == a[j] ? i < j : a[i] > a[j]; });
  Int m;
  cin >> m;
  vector<Int> ks(m), pos(m);
  for (Int i = 0, max_i = (m); i < max_i; i++) cin >> ks[i] >> pos[i];
  vector<Int> qs(m);
  for (Int i = 0, max_i = (m); i < max_i; i++) qs[i] = i;
  sort(begin(qs), end(qs), [&](Int i, Int j) { return ks[i] < ks[j]; });
  vector<Int> ans(m);
  BinaryIndexedTree bit(n);
  Int curk = 0;
  for (Int q = 0, max_i = (m); q < max_i; q++) {
    Int k = ks[qs[q]], p = pos[qs[q]];
    while (curk < k) {
      assert(bit[ord[curk]] == 0);
      bit.add(ord[curk], 1);
      curk++;
    }
    Int idx = bit.lower_bound(p);
    ans[qs[q]] = a[idx];
  }
  for (Int x : ans) cout << x << '\n';
}

```