#include <bits/stdc++.h>
using namespace std;
long double pi = 3.141592653589793238;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long n;
  cin >> n;
  vector<pair<long long, long long>> v1(n + 1);
  for (long long i = 1; i <= n; i++) cin >> v1[i].first;
  for (long long i = 1; i <= n; i++) cin >> v1[i].second;
  sort(v1.begin() + 1, v1.end());
  v1.push_back(make_pair(-1, -1));
  long long sum = 0, ans = 0, cnt = 0;
  priority_queue<long long> p;
  for (long long i = 1; i <= n; i++) {
    long long next = v1[i + 1].first;
    long long curr = v1[i].first;
    sum = sum + v1[i].second;
    cnt++;
    p.push(v1[i].second);
    while (cnt != 0 && curr != next) {
      long long f1 = p.top();
      p.pop();
      sum -= f1;
      ans += sum;
      cnt--;
      curr++;
    }
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
long double pi = 3.141592653589793238;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long n;
  cin >> n;
  vector<pair<long long, long long>> v1(n + 1);
  for (long long i = 1; i <= n; i++) cin >> v1[i].first;
  for (long long i = 1; i <= n; i++) cin >> v1[i].second;
  sort(v1.begin() + 1, v1.end());
  v1.push_back(make_pair(-1, -1));
  long long sum = 0, ans = 0, cnt = 0;
  priority_queue<long long> p;
  for (long long i = 1; i <= n; i++) {
    long long next = v1[i + 1].first;
    long long curr = v1[i].first;
    sum = sum + v1[i].second;
    cnt++;
    p.push(v1[i].second);
    while (cnt != 0 && curr != next) {
      long long f1 = p.top();
      p.pop();
      sum -= f1;
      ans += sum;
      cnt--;
      curr++;
    }
  }
  cout << ans << endl;
}

```