#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  std::vector<pair<int, int>> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].first;
  }
  for (int i = 0; i < n; i++) {
    cin >> v[i].second;
  }
  sort(v.begin(), v.end());
  long long int ans = 0;
  priority_queue<int> q;
  q.push(v[0].second);
  long long int sum = v[0].second;
  for (int i = 1; i < n; i++) {
    int d = v[i].first - v[i - 1].first;
    for (int j = 0; j < d; j++) {
      if (q.empty()) break;
      int temp = q.top();
      q.pop();
      sum -= temp;
      ans += sum;
    }
    sum += v[i].second;
    q.push(v[i].second);
  }
  for (int i = 0; i < n; i++) {
    if (q.empty()) break;
    int temp = q.top();
    q.pop();
    sum -= temp;
    ans += sum;
  }
  cout << ans;
}

### Prompt

Your challenge is to write a Cpp solution to 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;
  std::vector<pair<int, int>> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].first;
  }
  for (int i = 0; i < n; i++) {
    cin >> v[i].second;
  }
  sort(v.begin(), v.end());
  long long int ans = 0;
  priority_queue<int> q;
  q.push(v[0].second);
  long long int sum = v[0].second;
  for (int i = 1; i < n; i++) {
    int d = v[i].first - v[i - 1].first;
    for (int j = 0; j < d; j++) {
      if (q.empty()) break;
      int temp = q.top();
      q.pop();
      sum -= temp;
      ans += sum;
    }
    sum += v[i].second;
    q.push(v[i].second);
  }
  for (int i = 0; i < n; i++) {
    if (q.empty()) break;
    int temp = q.top();
    q.pop();
    sum -= temp;
    ans += sum;
  }
  cout << ans;
}

```