#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10, mod = 1e9 + 7, inf = 1e9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  pair<long long, long long> pai[n];
  for (int i = 0; i < n; i++) cin >> pai[i].first;
  for (int i = 0; i < n; i++) cin >> pai[i].second;
  sort(pai, pai + n);
  multiset<long long> s;
  long long ans = 0, a = 0, su = 0, i = 0;
  while (true) {
    if (s.size() == 0 && i == n) {
      cout << ans << "\n";
      return 0;
    } else if (s.size() == 0) {
      a = pai[i].first;
      su = 0;
    }
    for (i; (i < n && pai[i].first == a); i++) {
      s.insert(pai[i].second);
      su += pai[i].second;
    }
    su -= *--s.end();
    ans += su;
    s.erase(s.find(*--s.end()));
    a++;
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.

The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds. 

What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.

Input

The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).

The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).

The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).

Output

Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.

Examples

Input


5
3 7 9 7 8
5 2 5 7 5


Output


6


Input


5
1 2 3 4 5
1 1 1 1 1


Output


0

Note

In the first example, it is possible to find three publications of the second type, which will take 6 seconds.

In the second example, all news categories contain a different number of publications.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10, mod = 1e9 + 7, inf = 1e9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  pair<long long, long long> pai[n];
  for (int i = 0; i < n; i++) cin >> pai[i].first;
  for (int i = 0; i < n; i++) cin >> pai[i].second;
  sort(pai, pai + n);
  multiset<long long> s;
  long long ans = 0, a = 0, su = 0, i = 0;
  while (true) {
    if (s.size() == 0 && i == n) {
      cout << ans << "\n";
      return 0;
    } else if (s.size() == 0) {
      a = pai[i].first;
      su = 0;
    }
    for (i; (i < n && pai[i].first == a); i++) {
      s.insert(pai[i].second);
      su += pai[i].second;
    }
    su -= *--s.end();
    ans += su;
    s.erase(s.find(*--s.end()));
    a++;
  }
  return 0;
}

```