#include <bits/stdc++.h>
using namespace std;
const int M = 200200;
long long a[M], t[M];
pair<long long, long long> b[M];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> t[i];
    b[i] = make_pair(a[i], t[i]);
  }
  sort(b, b + n);
  long long sum = 0;
  long long ans = 0;
  multiset<int> rem;
  long long cur = 0;
  int i = 0;
  while (i < n || !rem.empty()) {
    if (rem.empty()) {
      cur = b[i].first;
    }
    while (i < n && cur == b[i].first) {
      rem.insert(-b[i].second);
      sum += b[i].second;
      ++i;
    }
    if (!rem.empty()) {
      long long mn = -*rem.begin();
      rem.erase(rem.begin());
      sum -= mn;
      ans += sum;
      ++cur;
    }
  }
  cout << ans << "\n";
}

### 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;
const int M = 200200;
long long a[M], t[M];
pair<long long, long long> b[M];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> t[i];
    b[i] = make_pair(a[i], t[i]);
  }
  sort(b, b + n);
  long long sum = 0;
  long long ans = 0;
  multiset<int> rem;
  long long cur = 0;
  int i = 0;
  while (i < n || !rem.empty()) {
    if (rem.empty()) {
      cur = b[i].first;
    }
    while (i < n && cur == b[i].first) {
      rem.insert(-b[i].second);
      sum += b[i].second;
      ++i;
    }
    if (!rem.empty()) {
      long long mn = -*rem.begin();
      rem.erase(rem.begin());
      sum -= mn;
      ans += sum;
      ++cur;
    }
  }
  cout << ans << "\n";
}

```