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;
struct node {
  long long x, t;
  bool operator<(const node &f) const { return t > f.t; }
} a[200005];
bool cmp(node x, node y) {
  if (x.x == y.x) return x.t > y.t;
  return x.x < y.x;
}
map<long long, int> m;
map<long long, long long> f;
map<long long, long long> cnt;
int main() {
  long long n;
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i].x);
    cnt[a[i].x]++;
    m[a[i].x]++;
  }
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i].t);
  long long ans = 0;
  sort(a + 1, a + 1 + n, cmp);
  priority_queue<long long> q;
  for (int i = 1; i <= n; i++) {
    long long sum = 0;
    sum += a[i].t;
    int j = i;
    long long now = a[i].x;
    q.push(a[i].t);
    while (!q.empty()) {
      while (a[i + 1].x == now && i + 1 <= n) {
        sum += a[i + 1].t;
        i++;
        q.push(a[i].t);
      }
      long long k = q.top();
      q.pop();
      ans += sum - k;
      sum -= k;
      now++;
    }
  }
  printf("%lld\n", ans);
}
