#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const int INF = 1e9;
const long long IINF = 1e18;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const char dir[4] = {'D', 'R', 'U', 'L'};
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &x : v) is >> x;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < v.size(); ++i) {
    os << v[i] << (i + 1 == v.size() ? "" : " ");
  }
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << '(' << p.first << ',' << p.second << ')';
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const map<T, U> &m) {
  os << '{';
  for (auto itr = m.begin(); itr != m.end(); ++itr) {
    os << '(' << itr->first << ',' << itr->second << ')';
    if (++itr != m.end()) os << ',';
    --itr;
  }
  os << '}';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &s) {
  os << '{';
  for (auto itr = s.begin(); itr != s.end(); ++itr) {
    os << *itr;
    if (++itr != s.end()) os << ',';
    --itr;
  }
  os << '}';
  return os;
}
void debug_out() { cerr << '\n'; }
template <class Head, class... Tail>
void debug_out(Head &&head, Tail &&...tail) {
  cerr << head;
  if (sizeof...(Tail) > 0) cerr << ", ";
  debug_out(move(tail)...);
}
template <typename T>
T gcd(T x, T y) {
  return y != 0 ? gcd(y, x % y) : x;
}
template <typename T>
T lcm(T x, T y) {
  return x / gcd(x, y) * y;
}
template <class T1, class T2>
inline bool chmin(T1 &a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T1, class T2>
inline bool chmax(T1 &a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
struct BinaryIndexedTree {
  vector<T> dat;
  BinaryIndexedTree(int n) : dat(n + 1, 0) {}
  void add(int i, T x) {
    if (i == 0) return;
    for (; i <= dat.size(); i += (i & -i)) dat[i] += x;
  }
  T sum(int i) {
    T res = 0;
    for (; i > 0; i -= (i & -i)) res += dat[i];
    return res;
  }
  T query(int l, int r) { return sum(r - 1) - sum(l - 1); }
  int lower_bound(T x) {
    if (x <= 0) return 0;
    int lb = 0, r = 1;
    while (r < dat.size()) r <<= 1;
    for (; r > 0; r >>= 1) {
      if (lb + r < dat.size() && dat[lb + r] < x) {
        x -= dat[lb + r];
        lb += r;
      }
    }
    return lb + 1;
  }
  void add0(int i, T x) { add(i + 1, x); }
  T sum0(int i) { return sum(i + 1); }
  T query0(int l, int r) { return sum(r) - sum(l); }
};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  cin >> a;
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int>> Q;
  vector<pair<int, int>> A;
  for (int i = 0; i < m; ++i) {
    int k, p;
    cin >> k >> p;
    Q.push_back({{k, p}, i});
  }
  sort((Q).begin(), (Q).end());
  for (int i = 0; i < n; ++i) A.emplace_back(a[i], -i);
  sort((A).begin(), (A).end());
  reverse((A).begin(), (A).end());
  BinaryIndexedTree<int> BIT(n);
  int now = 0;
  vector<int> ans(m);
  for (int i = 0; i < m; ++i) {
    int k = Q[i].first.first, p = Q[i].first.second;
    while (now < k) BIT.add0(-A[now++].second, 1);
    ans[Q[i].second] = a[BIT.lower_bound(p) - 1];
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << '\n';
}

### 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;
const long long MOD = 1e9 + 7;
const int INF = 1e9;
const long long IINF = 1e18;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const char dir[4] = {'D', 'R', 'U', 'L'};
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &x : v) is >> x;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < v.size(); ++i) {
    os << v[i] << (i + 1 == v.size() ? "" : " ");
  }
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << '(' << p.first << ',' << p.second << ')';
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const map<T, U> &m) {
  os << '{';
  for (auto itr = m.begin(); itr != m.end(); ++itr) {
    os << '(' << itr->first << ',' << itr->second << ')';
    if (++itr != m.end()) os << ',';
    --itr;
  }
  os << '}';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &s) {
  os << '{';
  for (auto itr = s.begin(); itr != s.end(); ++itr) {
    os << *itr;
    if (++itr != s.end()) os << ',';
    --itr;
  }
  os << '}';
  return os;
}
void debug_out() { cerr << '\n'; }
template <class Head, class... Tail>
void debug_out(Head &&head, Tail &&...tail) {
  cerr << head;
  if (sizeof...(Tail) > 0) cerr << ", ";
  debug_out(move(tail)...);
}
template <typename T>
T gcd(T x, T y) {
  return y != 0 ? gcd(y, x % y) : x;
}
template <typename T>
T lcm(T x, T y) {
  return x / gcd(x, y) * y;
}
template <class T1, class T2>
inline bool chmin(T1 &a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T1, class T2>
inline bool chmax(T1 &a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
struct BinaryIndexedTree {
  vector<T> dat;
  BinaryIndexedTree(int n) : dat(n + 1, 0) {}
  void add(int i, T x) {
    if (i == 0) return;
    for (; i <= dat.size(); i += (i & -i)) dat[i] += x;
  }
  T sum(int i) {
    T res = 0;
    for (; i > 0; i -= (i & -i)) res += dat[i];
    return res;
  }
  T query(int l, int r) { return sum(r - 1) - sum(l - 1); }
  int lower_bound(T x) {
    if (x <= 0) return 0;
    int lb = 0, r = 1;
    while (r < dat.size()) r <<= 1;
    for (; r > 0; r >>= 1) {
      if (lb + r < dat.size() && dat[lb + r] < x) {
        x -= dat[lb + r];
        lb += r;
      }
    }
    return lb + 1;
  }
  void add0(int i, T x) { add(i + 1, x); }
  T sum0(int i) { return sum(i + 1); }
  T query0(int l, int r) { return sum(r) - sum(l); }
};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  cin >> a;
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int>> Q;
  vector<pair<int, int>> A;
  for (int i = 0; i < m; ++i) {
    int k, p;
    cin >> k >> p;
    Q.push_back({{k, p}, i});
  }
  sort((Q).begin(), (Q).end());
  for (int i = 0; i < n; ++i) A.emplace_back(a[i], -i);
  sort((A).begin(), (A).end());
  reverse((A).begin(), (A).end());
  BinaryIndexedTree<int> BIT(n);
  int now = 0;
  vector<int> ans(m);
  for (int i = 0; i < m; ++i) {
    int k = Q[i].first.first, p = Q[i].first.second;
    while (now < k) BIT.add0(-A[now++].second, 1);
    ans[Q[i].second] = a[BIT.lower_bound(p) - 1];
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << '\n';
}

```