#include <bits/stdc++.h>
using namespace std;
double PI = 3.141592653589793238462643383279;
const double EPS = 1e-9;
const long long MOD = 1000000007;
const int inf = 1 << 30;
const long long linf = 1LL << 60;
template <typename T>
struct segtree {
  int n;
  vector<T> dat;
  T unit;
  void init(int n_, T unit_) {
    unit = unit_;
    n = 1;
    while (n < n_) n *= 2;
    dat.assign(2 * n - 1, unit);
  }
  void update(int k, T a) {
    k += n - 1;
    dat[k] = a;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = max(dat[k * 2 + 1], dat[k * 2 + 2]);
    }
  }
  T query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return unit;
    if (a <= l && r <= b)
      return dat[k];
    else {
      T vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
      T vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
      return max(vl, vr);
    }
  }
};
template <typename T>
struct segtree2 {
  int n;
  vector<T> dat;
  void init(int n_) {
    n = 1;
    while (n < n_) n *= 2;
    dat.assign(2 * n - 1, 0);
  }
  void update(int k, T a) {
    k += n - 1;
    dat[k] += a;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = dat[k * 2 + 1] + dat[k * 2 + 2];
    }
  }
  int query(int v, int k) {
    if (k >= n - 1) return k - (n - 1);
    if (dat[k * 2 + 1] >= v) return query(v, k * 2 + 1);
    return query(v - dat[k * 2 + 1], k * 2 + 2);
  }
};
vector<pair<int, int> > q[200000];
int n;
int a[200000];
int m;
int ans[200000];
segtree<pair<int, int> > seg;
segtree2<int> seg2;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < (int)n; i++) cin >> a[i];
  seg.init(n, pair<int, int>(-1, -1));
  seg2.init(n);
  for (int i = 0; i < (int)n; i++) {
    seg.update(i, pair<int, int>(a[i], -1 * i));
  }
  cin >> m;
  for (int i = 0; i < (int)m; i++) {
    int k, pos;
    cin >> k >> pos;
    q[k - 1].push_back(pair<int, int>(pos, i));
  }
  for (int i = 0; i < (int)n; i++) {
    pair<int, int> r = seg.query(0, seg.n, 0, 0, seg.n);
    r.second *= -1;
    seg.update(r.second, pair<int, int>(-1, -1));
    seg2.update(r.second, 1);
    for (int j = 0; j < (int)q[i].size(); j++) {
      int p = seg2.query(q[i][j].first, 0);
      ans[q[i][j].second] = a[p];
    }
  }
  for (int i = 0; i < (int)m; i++) cout << ans[i] << endl;
}

### Prompt

Please create a solution in Cpp 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;
double PI = 3.141592653589793238462643383279;
const double EPS = 1e-9;
const long long MOD = 1000000007;
const int inf = 1 << 30;
const long long linf = 1LL << 60;
template <typename T>
struct segtree {
  int n;
  vector<T> dat;
  T unit;
  void init(int n_, T unit_) {
    unit = unit_;
    n = 1;
    while (n < n_) n *= 2;
    dat.assign(2 * n - 1, unit);
  }
  void update(int k, T a) {
    k += n - 1;
    dat[k] = a;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = max(dat[k * 2 + 1], dat[k * 2 + 2]);
    }
  }
  T query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return unit;
    if (a <= l && r <= b)
      return dat[k];
    else {
      T vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
      T vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
      return max(vl, vr);
    }
  }
};
template <typename T>
struct segtree2 {
  int n;
  vector<T> dat;
  void init(int n_) {
    n = 1;
    while (n < n_) n *= 2;
    dat.assign(2 * n - 1, 0);
  }
  void update(int k, T a) {
    k += n - 1;
    dat[k] += a;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = dat[k * 2 + 1] + dat[k * 2 + 2];
    }
  }
  int query(int v, int k) {
    if (k >= n - 1) return k - (n - 1);
    if (dat[k * 2 + 1] >= v) return query(v, k * 2 + 1);
    return query(v - dat[k * 2 + 1], k * 2 + 2);
  }
};
vector<pair<int, int> > q[200000];
int n;
int a[200000];
int m;
int ans[200000];
segtree<pair<int, int> > seg;
segtree2<int> seg2;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < (int)n; i++) cin >> a[i];
  seg.init(n, pair<int, int>(-1, -1));
  seg2.init(n);
  for (int i = 0; i < (int)n; i++) {
    seg.update(i, pair<int, int>(a[i], -1 * i));
  }
  cin >> m;
  for (int i = 0; i < (int)m; i++) {
    int k, pos;
    cin >> k >> pos;
    q[k - 1].push_back(pair<int, int>(pos, i));
  }
  for (int i = 0; i < (int)n; i++) {
    pair<int, int> r = seg.query(0, seg.n, 0, 0, seg.n);
    r.second *= -1;
    seg.update(r.second, pair<int, int>(-1, -1));
    seg2.update(r.second, 1);
    for (int j = 0; j < (int)q[i].size(); j++) {
      int p = seg2.query(q[i][j].first, 0);
      ans[q[i][j].second] = a[p];
    }
  }
  for (int i = 0; i < (int)m; i++) cout << ans[i] << endl;
}

```