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>
long long n, sol, cnt, it = 1, sum;
std::pair<long long, long long> list[200005];
std::priority_queue<long long> pq;
int main() {
  std::cin >> n;
  for (int i = 1; i <= n; ++i) std::cin >> list[i].first;
  for (int i = 1; i <= n; ++i) std::cin >> list[i].second;
  std::sort(list + 1, list + n + 1);
  while (it <= n) {
    if (list[it].first == cnt) {
      pq.push(list[it].second);
      sum += list[it++].second;
      continue;
    } else if (pq.empty()) {
      cnt = list[it].first;
      pq.push(list[it].second);
      sum += list[it++].second;
    } else {
      sum -= pq.top();
      sol += sum;
      pq.pop();
      if (!pq.empty())
        ++cnt;
      else
        cnt = -1;
    }
  }
  while (!pq.empty()) {
    sum -= pq.top();
    sol += sum;
    pq.pop();
  }
  std::cout << sol;
  return 0;
}
