#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, x, z, res = 0, f, curr, ans = 0, y;
  cin >> n;
  int a[n];
  int b[n];
  map<int, int> m;
  unordered_map<int, vector<int> > u;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  for (int i = 0; i < n; i++) {
    m[a[i]]++;
    u[a[i]].push_back(b[i]);
  }
  map<int, int>::iterator it;
  priority_queue<int> pq;
  for (it = m.begin(); it != m.end(); it++) {
    x = (*it).first;
    y = (*it).second;
    if (y > 1 || !pq.empty()) {
      for (int i = 0; i < u[x].size(); i++) {
        pq.push(u[x][i]);
        ans += u[x][i];
      }
      ans -= pq.top();
      pq.pop();
      curr = x;
      while (!pq.empty()) {
        res += ans;
        curr++;
        if (u[curr].size())
          break;
        else {
          ans -= pq.top();
          pq.pop();
        }
      }
    }
  }
  cout << res << "\n";
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int n, x, z, res = 0, f, curr, ans = 0, y;
  cin >> n;
  int a[n];
  int b[n];
  map<int, int> m;
  unordered_map<int, vector<int> > u;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  for (int i = 0; i < n; i++) {
    m[a[i]]++;
    u[a[i]].push_back(b[i]);
  }
  map<int, int>::iterator it;
  priority_queue<int> pq;
  for (it = m.begin(); it != m.end(); it++) {
    x = (*it).first;
    y = (*it).second;
    if (y > 1 || !pq.empty()) {
      for (int i = 0; i < u[x].size(); i++) {
        pq.push(u[x][i]);
        ans += u[x][i];
      }
      ans -= pq.top();
      pq.pop();
      curr = x;
      while (!pq.empty()) {
        res += ans;
        curr++;
        if (u[curr].size())
          break;
        else {
          ans -= pq.top();
          pq.pop();
        }
      }
    }
  }
  cout << res << "\n";
}

```