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;
long long n, k, sum, cur, ans;
multiset<long long> second;
pair<long long, long long> a[200005];
int main() {
  cin >> n;
  for (k = 1; k <= n; k++) {
    cin >> a[k].first;
  }
  for (k = 1; k <= n; k++) {
    cin >> a[k].second;
  }
  sort(a + 1, a + n + 1);
  for (k = 1; k <= n + 1; k++) {
    if (a[k].first == cur) {
      second.insert(a[k].second);
      sum += a[k].second;
    } else {
      while (second.size() && cur != a[k].first) {
        sum -= *--second.end();
        second.erase(second.find(*--second.end()));
        ans += sum;
        cur++;
      }
      second.insert(a[k].second);
      sum += a[k].second;
      cur = a[k].first;
    }
    if (cur != a[k + 1].first) {
      sum -= (*--second.end());
      ans += sum;
      second.erase(second.find(*--second.end()));
      if (second.size()) {
        cur++;
      }
    }
  }
  cout << ans;
}
