#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; i++) cin >> a[i].first;
  for (int i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end());
  long long int ans = 0;
  multiset<int> s;
  s.insert(a[0].second);
  long long int sum = a[0].second;
  for (int i = 1; i < n; i++) {
    if (a[i].first == a[i - 1].first) {
      sum += a[i].second;
      s.insert(a[i].second);
    } else {
      if (!s.empty()) {
        sum = sum - *s.rbegin();
        s.erase(--s.end());
        ans = ans + sum;
        if (i < n) {
          int mn = a[i - 1].first + 1;
          while (!s.empty() && mn < a[i].first) {
            sum = sum - *s.rbegin();
            ans += sum;
            s.erase(--s.end());
            mn++;
          }
        }
        sum += a[i].second;
        s.insert(a[i].second);
      }
    }
  }
  while (!s.empty()) {
    sum = sum - *s.rbegin();
    s.erase(--s.end());
    ans = ans + sum;
  }
  cout << ans << endl;
}

### 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() {
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; i++) cin >> a[i].first;
  for (int i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end());
  long long int ans = 0;
  multiset<int> s;
  s.insert(a[0].second);
  long long int sum = a[0].second;
  for (int i = 1; i < n; i++) {
    if (a[i].first == a[i - 1].first) {
      sum += a[i].second;
      s.insert(a[i].second);
    } else {
      if (!s.empty()) {
        sum = sum - *s.rbegin();
        s.erase(--s.end());
        ans = ans + sum;
        if (i < n) {
          int mn = a[i - 1].first + 1;
          while (!s.empty() && mn < a[i].first) {
            sum = sum - *s.rbegin();
            ans += sum;
            s.erase(--s.end());
            mn++;
          }
        }
        sum += a[i].second;
        s.insert(a[i].second);
      }
    }
  }
  while (!s.empty()) {
    sum = sum - *s.rbegin();
    s.erase(--s.end());
    ans = ans + sum;
  }
  cout << ans << endl;
}

```