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

### 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;
signed main() {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  for (long long i = 0; i < n; i++) cin >> a[i].first;
  for (long long i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end());
  long long ans = 0;
  priority_queue<pair<long long, long long> > q;
  q.push({0, 0});
  for (long long i = 0; i < n; i++) {
    if (a[i].first != a[i - 1].first) {
      if (q.size() == 1) {
        q.pop();
        q.push({a[i].second, a[i].first});
        continue;
      }
      long long cur = a[i - 1].first;
      while (q.size() > 0 && cur < a[i].first) {
        pair<long long, long long> p = q.top();
        q.pop();
        ans += (cur - p.second) * p.first;
        cur++;
      }
      q.push({a[i].second, a[i].first});
    } else {
      q.push({a[i].second, a[i].first});
    }
  }
  long long cur = a[n - 1].first;
  while (q.size() > 0) {
    pair<long long, long long> p = q.top();
    q.pop();
    ans += (cur - p.second) * p.first;
    cur++;
  }
  cout << ans << "\n";
}

```