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;
using Graph = vector<vector<int>>;
const int INF = 2000000000;
const long long INF64 = 9223372036854775806ll;
const int mod = 1000000007ll;
const long long MOD = 1000000007LL;
int main() {
  int n;
  std::cin >> n;
  std::vector<pair<long long, long long>> a(n);
  for (int i = 0; i < (int)(n); i++) {
    std::cin >> a[i].first;
  }
  for (int i = 0; i < (int)(n); i++) {
    std::cin >> a[i].second;
  }
  sort((a).begin(), (a).end());
  long long ans = 0ll;
  priority_queue<long long> p;
  p.push(a[0].second);
  long long accumulate = a[0].second;
  for (int i = 1; i < (int)(n); i++) {
    for (int j = 0; j < (int)(a[i].first - a[i - 1].first); j++) {
      if (p.empty()) break;
      accumulate -= p.top();
      ans += accumulate;
      p.pop();
    }
    p.push(a[i].second);
    accumulate += a[i].second;
  }
  accumulate -= p.top();
  p.pop();
  while (!p.empty()) {
    ans += accumulate;
    accumulate -= p.top();
    p.pop();
  }
  std::cout << ans << std::endl;
}
