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.
#include <bits/stdc++.h>
using namespace std;
int n;
int a[500005];
int t[500005];
int book[500005];
int book_size;
int b[500005];
void disc(int *book, int &book_size) {
  sort(book + 1, book + 1 + book_size);
  book_size = (int)(unique(book + 1, book + 1 + book_size) - book - 1);
}
int get_rak(int x) {
  return (int)(lower_bound(book + 1, book + 1 + book_size, x) - book);
}
vector<int> G[500005];
priority_queue<int> pq;
long long ans;
long long sum;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    book[++book_size] = a[i];
  }
  book[++book_size] = 0x3f3f3f3f;
  disc(book, book_size);
  for (int i = 1; i <= n; ++i) b[i] = get_rak(a[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", t + i);
  for (int i = 1; i <= n; ++i) G[b[i]].push_back(t[i]);
  for (int i = 1; i < book_size; ++i) {
    for (auto t : G[i]) sum += t, pq.push(t);
    for (int j = book[i]; j < book[i + 1]; ++j) {
      if (pq.empty()) break;
      sum -= pq.top();
      pq.pop();
      ans += sum;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
