#include <bits/stdc++.h>
using namespace std;
struct Str {
  int ans, x, y, num;
};
int t[1000100];
int n, a[200200], q;
vector<pair<int, int> > b(200200);
Str query[200200];
bool comp(pair<int, int>& d, pair<int, int>& d1) {
  if (d.first != d1.first) {
    return (d.first < d1.first);
  } else {
    return (d.second > d1.second);
  }
}
bool comp1(Str& d, Str& d1) {
  if (d.x != d1.x) {
    return (d.x > d1.x);
  }
  return (d.y < d1.y);
}
bool comp2(Str& d, Str& d1) { return (d.num < d1.num); }
void build(int ver, int l, int r) {
  if (l == r) {
    t[ver] = 1;
    return;
  }
  int mid = l + (r - l) / 2;
  build(ver * 2, l, mid);
  build(ver * 2 + 1, mid + 1, r);
  t[ver] = t[ver * 2] + t[ver * 2 + 1];
}
void update(int ver, int l, int r, int pos) {
  if (l > pos || r < pos) {
    return;
  }
  if (l == r) {
    t[ver] = 0;
    return;
  }
  int mid = l + (r - l) / 2;
  update(ver * 2, l, mid, pos);
  update(ver * 2 + 1, mid + 1, r, pos);
  t[ver] = t[ver * 2] + t[ver * 2 + 1];
}
int getAns(int ver, int l, int r, int val) {
  if (l == r) {
    return l;
  }
  int mid = l + (r - l) / 2;
  if (t[ver * 2] >= val) {
    return getAns(ver * 2, l, mid, val);
  } else {
    return getAns(ver * 2 + 1, mid + 1, r, val - t[ver * 2]);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i].first = a[i];
    b[i].second = i;
  }
  sort(b.begin() + 1, b.begin() + n + 1, comp);
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> query[i].x >> query[i].y;
    query[i].num = i;
  }
  build(1, 1, n);
  sort(query + 1, query + q + 1, comp1);
  int cnt = n;
  for (int i = 1; i <= q; i++) {
    while (cnt != query[i].x) {
      update(1, 1, n, b[n - cnt + 1].second);
      cnt--;
    }
    query[i].ans = a[getAns(1, 1, n, query[i].y)];
  }
  sort(query + 1, query + q + 1, comp2);
  for (int i = 1; i <= q; i++) {
    cout << query[i].ans << "\n";
  }
}

### Prompt

Your task is to create 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;
struct Str {
  int ans, x, y, num;
};
int t[1000100];
int n, a[200200], q;
vector<pair<int, int> > b(200200);
Str query[200200];
bool comp(pair<int, int>& d, pair<int, int>& d1) {
  if (d.first != d1.first) {
    return (d.first < d1.first);
  } else {
    return (d.second > d1.second);
  }
}
bool comp1(Str& d, Str& d1) {
  if (d.x != d1.x) {
    return (d.x > d1.x);
  }
  return (d.y < d1.y);
}
bool comp2(Str& d, Str& d1) { return (d.num < d1.num); }
void build(int ver, int l, int r) {
  if (l == r) {
    t[ver] = 1;
    return;
  }
  int mid = l + (r - l) / 2;
  build(ver * 2, l, mid);
  build(ver * 2 + 1, mid + 1, r);
  t[ver] = t[ver * 2] + t[ver * 2 + 1];
}
void update(int ver, int l, int r, int pos) {
  if (l > pos || r < pos) {
    return;
  }
  if (l == r) {
    t[ver] = 0;
    return;
  }
  int mid = l + (r - l) / 2;
  update(ver * 2, l, mid, pos);
  update(ver * 2 + 1, mid + 1, r, pos);
  t[ver] = t[ver * 2] + t[ver * 2 + 1];
}
int getAns(int ver, int l, int r, int val) {
  if (l == r) {
    return l;
  }
  int mid = l + (r - l) / 2;
  if (t[ver * 2] >= val) {
    return getAns(ver * 2, l, mid, val);
  } else {
    return getAns(ver * 2 + 1, mid + 1, r, val - t[ver * 2]);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i].first = a[i];
    b[i].second = i;
  }
  sort(b.begin() + 1, b.begin() + n + 1, comp);
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> query[i].x >> query[i].y;
    query[i].num = i;
  }
  build(1, 1, n);
  sort(query + 1, query + q + 1, comp1);
  int cnt = n;
  for (int i = 1; i <= q; i++) {
    while (cnt != query[i].x) {
      update(1, 1, n, b[n - cnt + 1].second);
      cnt--;
    }
    query[i].ans = a[getAns(1, 1, n, query[i].y)];
  }
  sort(query + 1, query + q + 1, comp2);
  for (int i = 1; i <= q; i++) {
    cout << query[i].ans << "\n";
  }
}

```