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;
long long res[300000];
int cd;
map<long long, long long> ok;
pair<long long, long long> a[300000];
priority_queue<long long> pq;
int main() {
  int t = 1;
  while (t--) {
    cin >> cd;
    for (int i = 1; i <= cd; i++) {
      cin >> a[i].first;
      ok[a[i].first] = 1;
    }
    for (int i = 1; i <= cd; i++) {
      cin >> a[i].second;
    }
    sort(a + 1, a + 1 + cd);
    cd++;
    a[cd].first = 1e17;
    long long kq = 0;
    long long cur = 0;
    int i = 1;
    while (i < cd) {
      pq.push(a[i].second);
      cur += a[i].second;
      while (a[i + 1].first == a[i].first) {
        i++;
        pq.push(a[i].second);
        cur += a[i].second;
      }
      cur -= pq.top();
      pq.pop();
      for (long long e = a[i].first + 1; e <= 1000000000000; e++) {
        if (pq.size() == 0) {
          break;
        }
        kq += cur;
        if (ok[e] == 1) {
          break;
        }
        cur -= pq.top();
        pq.pop();
      }
      i++;
    }
    cout << kq << '\n';
  }
}
