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;
namespace yspm {
inline long long read() {
  long long res = 0, f = 1;
  char k;
  while (!isdigit(k = getchar()))
    if (k == '-') f = -1;
  while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
  return res * f;
}
const long long N = 2e5 + 10;
long long a[N], val[N], n;
struct node {
  long long first, second;
  bool operator<(const node &x) const {
    return first < x.first || (first == x.first && second < x.second);
  }
} x[N];
priority_queue<long long> q;
long long ans;
signed main() {
  n = read();
  for (long long i = 1; i <= n; ++i) x[i].first = read();
  for (long long i = 1; i <= n; ++i) x[i].second = read();
  sort(x + 1, x + n + 1);
  x[n + 1].first = 0x3f3f3f3f3f3f3f3f;
  long long now = 0, sum = 0;
  for (long long i = 1; i <= n; ++i) {
    q.push(x[i].second);
    sum += x[i].second;
    now = x[i].first;
    while (q.size() && now < x[i + 1].first) {
      long long tmp = q.top();
      q.pop();
      sum -= tmp;
      ans += sum, now++;
    }
  }
  cout << ans << endl;
  return 0;
}
}  // namespace yspm
signed main() { return yspm::main(); }
