#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
inline void chmin(A& a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A& a, B b) {
  if (a < b) a = b;
}
template <class A, class B>
ostream& operator<<(ostream& ost, const pair<A, B>& p) {
  ost << "{" << p.first << "," << p.second << "}";
  return ost;
}
template <class T>
ostream& operator<<(ostream& ost, const vector<T>& v) {
  ost << "{";
  for (long long i = 0; i < v.size(); i++) {
    if (i) ost << ",";
    ost << v[i];
  }
  ost << "}";
  return ost;
}
template <class T>
struct BinaryIndexedTree {
  long long n;
  vector<T> dat;
  BinaryIndexedTree(long long n = 0) : n(n), dat(n + 1) {}
  void add(long long k, T x) {
    for (k++; k <= n; k += k & -k) dat[k] += x;
  }
  T sum(long long k) {
    T ret{};
    for (k++; k; k -= k & -k) ret += dat[k];
    return ret;
  }
};
long long N;
long long A[222222];
long long K[222222], P[222222];
long long Q;
signed main() {
  scanf("%lld", &N);
  for (long long i = 0; i < (N); i++) scanf("%lld", &A[i]);
  vector<long long> ord(N);
  iota((ord).begin(), (ord).end(), 0);
  sort((ord).begin(), (ord).end(), [&](long long i, long long j) {
    if (A[i] == A[j]) return i < j;
    return A[i] > A[j];
  });
  scanf("%lld", &Q);
  vector<vector<long long> > bucket(N);
  for (long long i = 0; i < (Q); i++) {
    scanf("%lld%lld", &K[i], &P[i]);
    K[i]--;
    P[i]--;
    bucket[K[i]].push_back(i);
  }
  BinaryIndexedTree<long long> bit(N + 114);
  vector<long long> ans(Q);
  for (long long i = 0; i < (N); i++) {
    bit.add(ord[i], 1);
    for (auto j : bucket[i]) {
      long long lb = 0, ub = N;
      while (ub - lb > 1) {
        long long mid = (ub + lb) / 2;
        if (bit.sum(mid - 1) <= P[j])
          lb = mid;
        else
          ub = mid;
      }
      ans[j] = A[lb];
    }
  }
  for (long long i = 0; i < (Q); i++) printf("%lld\n", ans[i]);
  return 0;
}

### Prompt

Please formulate 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;
template <typename A, typename B>
inline void chmin(A& a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A& a, B b) {
  if (a < b) a = b;
}
template <class A, class B>
ostream& operator<<(ostream& ost, const pair<A, B>& p) {
  ost << "{" << p.first << "," << p.second << "}";
  return ost;
}
template <class T>
ostream& operator<<(ostream& ost, const vector<T>& v) {
  ost << "{";
  for (long long i = 0; i < v.size(); i++) {
    if (i) ost << ",";
    ost << v[i];
  }
  ost << "}";
  return ost;
}
template <class T>
struct BinaryIndexedTree {
  long long n;
  vector<T> dat;
  BinaryIndexedTree(long long n = 0) : n(n), dat(n + 1) {}
  void add(long long k, T x) {
    for (k++; k <= n; k += k & -k) dat[k] += x;
  }
  T sum(long long k) {
    T ret{};
    for (k++; k; k -= k & -k) ret += dat[k];
    return ret;
  }
};
long long N;
long long A[222222];
long long K[222222], P[222222];
long long Q;
signed main() {
  scanf("%lld", &N);
  for (long long i = 0; i < (N); i++) scanf("%lld", &A[i]);
  vector<long long> ord(N);
  iota((ord).begin(), (ord).end(), 0);
  sort((ord).begin(), (ord).end(), [&](long long i, long long j) {
    if (A[i] == A[j]) return i < j;
    return A[i] > A[j];
  });
  scanf("%lld", &Q);
  vector<vector<long long> > bucket(N);
  for (long long i = 0; i < (Q); i++) {
    scanf("%lld%lld", &K[i], &P[i]);
    K[i]--;
    P[i]--;
    bucket[K[i]].push_back(i);
  }
  BinaryIndexedTree<long long> bit(N + 114);
  vector<long long> ans(Q);
  for (long long i = 0; i < (N); i++) {
    bit.add(ord[i], 1);
    for (auto j : bucket[i]) {
      long long lb = 0, ub = N;
      while (ub - lb > 1) {
        long long mid = (ub + lb) / 2;
        if (bit.sum(mid - 1) <= P[j])
          lb = mid;
        else
          ub = mid;
      }
      ans[j] = A[lb];
    }
  }
  for (long long i = 0; i < (Q); i++) printf("%lld\n", ans[i]);
  return 0;
}

```