#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  pair<int, int> A[n];
  for (int i = 0; i < n; i++) {
    cin >> A[i].first;
  }
  for (int i = 0; i < n; i++) {
    cin >> A[i].second;
  }
  sort(A, A + n);
  multiset<int> vaibhav;
  multiset<int>::iterator it;
  long long int sum = 0, num = 0;
  num += A[0].second;
  vaibhav.insert(A[0].second);
  for (int i = 1; i < n; i++) {
    if (A[i].first > A[i - 1].first) {
      int count = 0;
      it = vaibhav.end();
      it--;
      int r = *it;
      num -= *it;
      vaibhav.erase(it);
      while (!vaibhav.empty() && count != A[i].first - A[i - 1].first - 1) {
        count++;
        sum += num;
        it = vaibhav.end();
        it--;
        num -= *it;
        vaibhav.erase(it);
      }
      sum += num;
      vaibhav.insert(min((int)1e9, A[i].second));
      num += min(A[i].second, (int)1e9);
    } else {
      vaibhav.insert(A[i].second);
      num += A[i].second;
    }
  }
  int count = 0;
  it = vaibhav.end();
  it--;
  num -= *it;
  vaibhav.erase(it);
  while (!vaibhav.empty()) {
    count++;
    sum += num;
    it = vaibhav.end();
    it--;
    num -= *it;
    vaibhav.erase(it);
  }
  cout << sum;
}
int main() {
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

### 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;
void solve() {
  int n;
  cin >> n;
  pair<int, int> A[n];
  for (int i = 0; i < n; i++) {
    cin >> A[i].first;
  }
  for (int i = 0; i < n; i++) {
    cin >> A[i].second;
  }
  sort(A, A + n);
  multiset<int> vaibhav;
  multiset<int>::iterator it;
  long long int sum = 0, num = 0;
  num += A[0].second;
  vaibhav.insert(A[0].second);
  for (int i = 1; i < n; i++) {
    if (A[i].first > A[i - 1].first) {
      int count = 0;
      it = vaibhav.end();
      it--;
      int r = *it;
      num -= *it;
      vaibhav.erase(it);
      while (!vaibhav.empty() && count != A[i].first - A[i - 1].first - 1) {
        count++;
        sum += num;
        it = vaibhav.end();
        it--;
        num -= *it;
        vaibhav.erase(it);
      }
      sum += num;
      vaibhav.insert(min((int)1e9, A[i].second));
      num += min(A[i].second, (int)1e9);
    } else {
      vaibhav.insert(A[i].second);
      num += A[i].second;
    }
  }
  int count = 0;
  it = vaibhav.end();
  it--;
  num -= *it;
  vaibhav.erase(it);
  while (!vaibhav.empty()) {
    count++;
    sum += num;
    it = vaibhav.end();
    it--;
    num -= *it;
    vaibhav.erase(it);
  }
  cout << sum;
}
int main() {
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

```