#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
priority_queue<int> q;
int n;
long long ans;
struct rec {
  int a, t;
  inline bool operator<(const rec &w) const { return a < w.a; }
} a[N];
inline int read() {
  int f = 0, fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + c - 48;
    c = getchar();
  }
  return f * fu;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i].a = read();
  for (int i = 1; i <= n; i++) a[i].t = read();
  sort(a + 1, a + 1 + n);
  int now = 1, num = 0;
  long long tmp = 0;
  while (!q.empty() || now <= n) {
    num++;
    if (q.empty()) {
      num = a[now].a;
      while (now <= n && num == a[now].a)
        tmp += a[now].t, q.push(a[now].t), now++;
    } else
      while (now <= n && num == a[now].a)
        tmp += a[now].t, q.push(a[now].t), now++;
    tmp -= q.top(), q.pop();
    ans += tmp;
  }
  printf("%lld\n", ans);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
priority_queue<int> q;
int n;
long long ans;
struct rec {
  int a, t;
  inline bool operator<(const rec &w) const { return a < w.a; }
} a[N];
inline int read() {
  int f = 0, fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + c - 48;
    c = getchar();
  }
  return f * fu;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i].a = read();
  for (int i = 1; i <= n; i++) a[i].t = read();
  sort(a + 1, a + 1 + n);
  int now = 1, num = 0;
  long long tmp = 0;
  while (!q.empty() || now <= n) {
    num++;
    if (q.empty()) {
      num = a[now].a;
      while (now <= n && num == a[now].a)
        tmp += a[now].t, q.push(a[now].t), now++;
    } else
      while (now <= n && num == a[now].a)
        tmp += a[now].t, q.push(a[now].t), now++;
    tmp -= q.top(), q.pop();
    ans += tmp;
  }
  printf("%lld\n", ans);
}

```