#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n;
  cin >> n;
  vector<long long> pub(n);
  vector<long long> t(n);
  for (int i = 0; i < n; i++) {
    cin >> pub[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  map<long long, multiset<long long>> mp;
  for (int i = 0; i < n; i++) {
    mp[pub[i]].insert(t[i]);
  }
  long long ans = 0;
  auto it = mp.begin();
  multiset<pair<long long, long long>, greater<pair<long long, long long>>> acc;
  long long last = (it->first) - 1;
  while (it != mp.end()) {
    auto see = acc.begin();
    while (see != acc.end()) {
      if (last + 1 < it->first) {
        last++;
        ans += (last - (see->second)) * (see->first);
        auto to_erase = see++;
        acc.erase(to_erase);
      } else {
        break;
      }
    }
    for (auto temp = (it->second).begin(); temp != (it->second).end(); temp++) {
      acc.insert({*temp, it->first});
    }
    last = it->first - 1;
    it++;
  }
  auto see = acc.begin();
  while (see != acc.end()) {
    last++;
    ans += (last - see->second) * (see->first);
    see++;
  }
  cout << ans << endl;
  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;
int main() {
  long long n;
  cin >> n;
  vector<long long> pub(n);
  vector<long long> t(n);
  for (int i = 0; i < n; i++) {
    cin >> pub[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  map<long long, multiset<long long>> mp;
  for (int i = 0; i < n; i++) {
    mp[pub[i]].insert(t[i]);
  }
  long long ans = 0;
  auto it = mp.begin();
  multiset<pair<long long, long long>, greater<pair<long long, long long>>> acc;
  long long last = (it->first) - 1;
  while (it != mp.end()) {
    auto see = acc.begin();
    while (see != acc.end()) {
      if (last + 1 < it->first) {
        last++;
        ans += (last - (see->second)) * (see->first);
        auto to_erase = see++;
        acc.erase(to_erase);
      } else {
        break;
      }
    }
    for (auto temp = (it->second).begin(); temp != (it->second).end(); temp++) {
      acc.insert({*temp, it->first});
    }
    last = it->first - 1;
    it++;
  }
  auto see = acc.begin();
  while (see != acc.end()) {
    last++;
    ans += (last - see->second) * (see->first);
    see++;
  }
  cout << ans << endl;
  return 0;
}

```