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>
const int maxn = 2e5 + 5;
using std::priority_queue;
template <class T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 1) + (x << 3) + (s & 15);
    s = getchar();
  }
  x *= f;
}
template <class T>
inline T min(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline T max(T a, T b) {
  return a > b ? a : b;
}
template <class T>
inline T fabs(T a) {
  return a > 0 ? a : -a;
}
struct node {
  long long a, t;
} a[maxn];
inline bool cmp(node a, node b) { return a.a < b.a; }
long long n, ans, s, sum;
priority_queue<long long> q;
inline void query(long long l) {
  for (long long i = 1; i <= l; i++) {
    if (q.empty()) break;
    s = q.top();
    sum -= s;
    q.pop();
    ans += sum;
  }
}
inline void push(long long x) {
  sum += a[x].t;
  q.push(a[x].t);
}
signed main() {
  read(n);
  for (long long i = 1; i <= n; i++) read(a[i].a);
  for (long long i = 1; i <= n; i++) read(a[i].t);
  std::sort(a + 1, a + n + 1, cmp);
  for (long long i = 1; i <= n; i++) {
    query(a[i].a - a[i - 1].a);
    push(i);
  }
  query(n);
  printf("%lld", ans);
}
