#include <bits/stdc++.h>
using namespace std;
signed main() {
  long long n;
  cin >> n;
  priority_queue<long long> p;
  vector<vector<long long>> a(n, vector<long long>(2));
  for (long long i = 0; i < n; i++) cin >> a[i][0];
  for (long long i = 0; i < n; i++) cin >> a[i][1];
  sort(a.begin(), a.end());
  long long i = 0, cur = 1, sum = 0, ans = 0;
  while (i < n || p.size()) {
    while (i < n && a[i][0] == cur) {
      sum += a[i][1];
      p.push(a[i][1]);
      i += 1;
    }
    if (p.size() == 0) {
      cur = a[i][0];
    } else {
      sum -= p.top();
      p.pop();
      ans += sum;
      cur += 1;
    }
  }
  cout << ans;
}

### Prompt

Your task is to create 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;
signed main() {
  long long n;
  cin >> n;
  priority_queue<long long> p;
  vector<vector<long long>> a(n, vector<long long>(2));
  for (long long i = 0; i < n; i++) cin >> a[i][0];
  for (long long i = 0; i < n; i++) cin >> a[i][1];
  sort(a.begin(), a.end());
  long long i = 0, cur = 1, sum = 0, ans = 0;
  while (i < n || p.size()) {
    while (i < n && a[i][0] == cur) {
      sum += a[i][1];
      p.push(a[i][1]);
      i += 1;
    }
    if (p.size() == 0) {
      cur = a[i][0];
    } else {
      sum -= p.top();
      p.pop();
      ans += sum;
      cur += 1;
    }
  }
  cout << ans;
}

```