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;
const long long maxn = 2e5 + 5;
inline long long read() {
  char c = getchar();
  long long t = 0, f = 1;
  while ((!isdigit(c)) && (c != EOF)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ((isdigit(c)) && (c != EOF)) {
    t = (t << 3) + (t << 1) + (c ^ 48);
    c = getchar();
  }
  return t * f;
}
long long n;
struct node {
  long long a, t;
} a[maxn];
priority_queue<long long> q;
bool cmp(node a, node b) { return a.a == b.a ? a.t < b.t : a.a < b.a; }
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i].a = read();
  for (long long i = 1; i <= n; i++) a[i].t = read();
  sort(a + 1, a + 1 + n, cmp);
  long long now = 0, ans = 0, sum = 0;
  for (long long i = 1; i <= n; i++) {
    long long cnt = 0;
    while ((!q.empty()) && (now + cnt < a[i].a)) {
      long long u = q.top();
      q.pop();
      ans = ans + u * cnt;
      sum -= u;
      cnt++;
    }
    ans = ans + sum * cnt;
    while (a[i + 1].a == a[i].a) {
      now = a[i].a;
      q.push(a[i].t);
      sum += a[i].t;
      i++;
    }
    now = a[i].a;
    sum += a[i].t;
    q.push(a[i].t);
  }
  long long cnt = 0;
  while (!q.empty()) {
    long long u = q.top();
    q.pop();
    ans = ans + u * cnt;
    cnt++;
  }
  printf("%lld\n", ans);
  return 0;
}
