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;
pair<long long, long long> a[200010];
priority_queue<long long> pq;
int main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i].first);
  }
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i].second);
  }
  a[n + 1].first = 1e18;
  sort(a + 1, a + 1 + n);
  long long ans = 0;
  long long now = 0;
  for (long long i = 1; i <= n; i++) {
    pq.push(a[i].second);
    if (a[i].first == a[i + 1].first) {
      now += a[i].second;
    } else {
      long long diff = a[i + 1].first - a[i].first;
      now += a[i].second;
      for (long long i = 1; i <= diff; i++) {
        if (pq.empty()) break;
        now -= pq.top();
        ans += now;
        pq.pop();
      }
    }
  }
  cout << ans << endl;
  return 0;
}
