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;
const int N = 2e5 + 5;
int n;
long long ans;
pair<int, int> p[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i].first;
  for (int i = 0; i < n; i++) cin >> p[i].second;
  sort(p, p + n);
  int pos = 0;
  set<pair<int, int> > s;
  long long x = -1;
  while (true) {
    while (!s.empty()) {
      pair<int, int> v = (*s.rbegin());
      ans += v.first * (x - p[v.second].first);
      s.erase(v);
      x++;
      while (pos < n && p[pos].first == x) {
        s.insert({p[pos].second, pos});
        pos++;
      }
    }
    if (pos == n) break;
    x = p[pos].first;
    while (pos < n && p[pos].first == x) {
      s.insert({p[pos].second, pos});
      pos++;
    }
  }
  cout << ans << "\n";
}
