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 = 200002;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<pair<int, int> > b(n);
  unsigned long long ans = 0;
  for (int i = 0; i < n; i++) cin >> b[i].first;
  for (int i = 0; i < n; i++) cin >> b[i].second;
  sort(b.begin(), b.end());
  int x = b[0].first;
  multiset<int> s;
  unsigned long long su = 0;
  for (int i = 0; i < n; i++) {
    if (b[i].first != x) {
      for (int j = x; j < b[i].first; j++) {
        if (s.empty()) break;
        int r = (*s.rbegin());
        s.erase(s.lower_bound(r));
        su -= r;
        ans += su;
      }
      x = b[i].first;
    }
    s.insert(b[i].second), su += b[i].second;
  }
  while (!s.empty()) {
    int r = (*s.rbegin());
    s.erase(s.lower_bound(r));
    su -= r;
    ans += su;
  }
  cout << ans << endl;
}
