#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string& s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
struct BIT {
  vector<int> bit;
  int n;
  BIT(int _n) {
    n = _n;
    bit = vector<int>(n + 1, 0);
  }
  void add(int x) {
    while (x <= n) {
      bit[x]++;
      x += (x & -x);
    }
  }
  int query(int x) {
    int ans = 0;
    while (x >= 1) {
      ans += bit[x];
      x -= (x & -x);
    }
    return ans;
  }
  int findAtRank(int k) {
    int lo = 1;
    int hi = n;
    int ans = -1;
    while (lo <= hi) {
      int mid = (lo + hi) / 2;
      if (query(mid) >= k) {
        ans = mid;
        hi = mid - 1;
      } else {
        lo = mid + 1;
      }
    }
    assert(ans != -1);
    return ans;
  }
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int& x : a) {
    cin >> x;
  }
  vector<vector<int> > queries(n + 1);
  vector<pair<int, int> > query_data;
  int Q;
  cin >> Q;
  while (Q--) {
    int k, pos;
    cin >> k >> pos;
    queries[k].push_back(pos);
    query_data.push_back({k, pos});
  }
  map<int, vector<int> > positions;
  for (int i = 0; i < n; i++) {
    positions[a[i]].push_back(i + 1);
  }
  set<int, greater<int> > st;
  for (int x : a) {
    st.insert(x);
  }
  BIT bit(n + 1);
  map<pair<int, int>, int> answer;
  int k = 0;
  for (int x : st) {
    for (int pos : positions[x]) {
      bit.add(pos);
      k++;
      if (queries[k].size() > 0) {
        for (int pos : queries[k]) {
          int ans_position = bit.findAtRank(pos);
          answer[{k, pos}] = a[ans_position - 1];
        }
      }
    }
  }
  for (auto it : query_data) {
    cout << answer[it] << "\n";
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string& s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
struct BIT {
  vector<int> bit;
  int n;
  BIT(int _n) {
    n = _n;
    bit = vector<int>(n + 1, 0);
  }
  void add(int x) {
    while (x <= n) {
      bit[x]++;
      x += (x & -x);
    }
  }
  int query(int x) {
    int ans = 0;
    while (x >= 1) {
      ans += bit[x];
      x -= (x & -x);
    }
    return ans;
  }
  int findAtRank(int k) {
    int lo = 1;
    int hi = n;
    int ans = -1;
    while (lo <= hi) {
      int mid = (lo + hi) / 2;
      if (query(mid) >= k) {
        ans = mid;
        hi = mid - 1;
      } else {
        lo = mid + 1;
      }
    }
    assert(ans != -1);
    return ans;
  }
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int& x : a) {
    cin >> x;
  }
  vector<vector<int> > queries(n + 1);
  vector<pair<int, int> > query_data;
  int Q;
  cin >> Q;
  while (Q--) {
    int k, pos;
    cin >> k >> pos;
    queries[k].push_back(pos);
    query_data.push_back({k, pos});
  }
  map<int, vector<int> > positions;
  for (int i = 0; i < n; i++) {
    positions[a[i]].push_back(i + 1);
  }
  set<int, greater<int> > st;
  for (int x : a) {
    st.insert(x);
  }
  BIT bit(n + 1);
  map<pair<int, int>, int> answer;
  int k = 0;
  for (int x : st) {
    for (int pos : positions[x]) {
      bit.add(pos);
      k++;
      if (queries[k].size() > 0) {
        for (int pos : queries[k]) {
          int ans_position = bit.findAtRank(pos);
          answer[{k, pos}] = a[ans_position - 1];
        }
      }
    }
  }
  for (auto it : query_data) {
    cout << answer[it] << "\n";
  }
  return 0;
}

```