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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int N;
  cin >> N;
  vector<pair<int, int> > A(N);
  for (int i = 0; i < N; i++) cin >> A[i].first;
  for (int i = 0; i < N; i++) cin >> A[i].second;
  sort((A).begin(), (A).end());
  int l = 0;
  long long res = 0;
  while (l < N) {
    int fin = A[l].first;
    int r = l + 1;
    while (r < N && A[r].first <= fin) {
      fin++;
      r++;
    }
    int idx = l;
    long long sum = 0;
    priority_queue<long long> pq;
    for (int i = A[l].first; i < fin + 1; i++) {
      while (idx < N && A[idx].first <= i) {
        sum += A[idx].second;
        pq.push(A[idx].second);
        idx++;
      }
      long long maxi = pq.top();
      pq.pop();
      sum -= maxi;
      res += sum;
    }
    l = r;
  }
  cout << res << '\n';
  return 0;
}
