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 + 123;
pair<int, int> a[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int val;
    scanf("%d", &val);
    a[i].first = val;
  }
  for (int i = 1; i <= n; ++i) {
    int cost;
    scanf("%d", &cost);
    a[i].second = cost;
  }
  sort(a + 1, a + 1 + n);
  long long ans = 0;
  priority_queue<pair<int, int> > q;
  q.push(make_pair(a[1].second, a[1].first));
  int cur = a[1].first;
  for (int i = 2; i <= n; ++i) {
    while (cur < a[i].first && !q.empty()) {
      pair<int, int> b = q.top();
      q.pop();
      ans += 1LL * b.first * (cur - b.second);
      ++cur;
    }
    cur = a[i].first;
    q.push(make_pair(a[i].second, a[i].first));
  }
  while (!q.empty()) {
    pair<int, int> b = q.top();
    q.pop();
    ans += 1LL * b.first * (cur - b.second);
    ++cur;
  }
  printf("%lld", ans);
  return 0;
}
