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;
inline int read() {
  int n = 0, f = 1, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = n * 10 + ch - '0';
    ch = getchar();
  }
  return n * f;
}
struct sth {
  int a, t;
} p[200005];
bool bi(struct sth x, struct sth y) {
  if (x.t != y.t) return x.t > y.t;
  return x.a < y.a;
}
map<int, int> fa;
int findf(int n) {
  if (fa[n] == 0) return n;
  return fa[n] = findf(fa[n]);
}
void merge(int x, int y) {
  x = findf(x);
  y = findf(y);
  if (x != y) fa[x] = y;
}
int main() {
  int n;
  n = read();
  for (int i = 1; i <= n; i++) p[i].a = read();
  for (int i = 1; i <= n; i++) p[i].t = read();
  sort(p + 1, p + n + 1, bi);
  long long ans = 0;
  int x;
  for (int i = 1; i <= n; i++) {
    x = findf(p[i].a);
    ans += 1LL * p[i].t * (x - p[i].a);
    merge(x, x + 1);
  }
  printf("%lld\n", ans);
  return 0;
}
