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;
pair<int, int> a[200200];
int n;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i].first;
  for (int i = 1; i <= n; i++) cin >> a[i].second;
  sort(a + 1, a + n + 1);
  int cur = a[1].first - 1;
  priority_queue<pair<int, int> > pq;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int foo = a[i].first;
    int t = min((int)pq.size(), foo - cur - 1);
    for (int j = 1; j <= t; j++) {
      int pos = cur + j;
      ans += 1ll * pq.top().first * (pos - pq.top().second);
      pq.pop();
    }
    pq.push({a[i].second, a[i].first});
    while (a[i + 1].first == a[i].first) {
      pq.push({a[i + 1].second, a[i + 1].first});
      i++;
    }
    ans += 1ll * pq.top().first * (foo - pq.top().second);
    pq.pop();
    cur = foo;
  }
  int t = (int)pq.size();
  for (int i = 1; i <= t; i++) {
    int pos = cur + i;
    ans += 1ll * pq.top().first * (pos - pq.top().second);
    pq.pop();
  }
  cout << ans << '\n';
}
