#include <bits/stdc++.h>
using namespace std;
struct Cate {
  long long a, t;
  bool operator<(const Cate &rhs) const {
    return a < rhs.a || (a == rhs.a && t < rhs.t);
  }
} category[200005];
long long n;
priority_queue<long long> p;
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) scanf("%lld", &category[i].a);
  for (long long i = 0; i < n; i++) scanf("%lld", &category[i].t);
  sort(category, category + n);
  long long cnt = 0;
  long long a = category[0].a;
  long long now = 0;
  long long sum = 0;
  while (true) {
    while (cnt < n && category[cnt].a == a) {
      p.push(category[cnt].t);
      now += category[cnt].t;
      cnt++;
    }
    now -= p.top();
    p.pop();
    sum += now;
    if (cnt == n && p.empty()) break;
    if (!p.empty())
      a++;
    else
      a = category[cnt].a;
  }
  cout << sum << endl;
}

### 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;
struct Cate {
  long long a, t;
  bool operator<(const Cate &rhs) const {
    return a < rhs.a || (a == rhs.a && t < rhs.t);
  }
} category[200005];
long long n;
priority_queue<long long> p;
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) scanf("%lld", &category[i].a);
  for (long long i = 0; i < n; i++) scanf("%lld", &category[i].t);
  sort(category, category + n);
  long long cnt = 0;
  long long a = category[0].a;
  long long now = 0;
  long long sum = 0;
  while (true) {
    while (cnt < n && category[cnt].a == a) {
      p.push(category[cnt].t);
      now += category[cnt].t;
      cnt++;
    }
    now -= p.top();
    p.pop();
    sum += now;
    if (cnt == n && p.empty()) break;
    if (!p.empty())
      a++;
    else
      a = category[cnt].a;
  }
  cout << sum << endl;
}

```