#include <bits/stdc++.h>
using namespace std;
struct Element {
  int value, cost;
  inline bool operator<(const Element &x) const { return value < x.value; }
};
Element a[200005];
priority_queue<int> heap;
int n, pnt, now, tmp;
long long ans, nowans;
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i].value);
  }
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i].cost);
  }
  sort(a + 1, a + n + 1);
  while (pnt < n || heap.empty() == false) {
    tmp++;
    if (heap.empty() == true) {
      tmp = a[pnt + 1].value;
    }
    while (pnt < n && a[pnt + 1].value == tmp) {
      pnt++;
      nowans += a[pnt].cost;
      heap.push(a[pnt].cost);
    }
    nowans -= heap.top();
    heap.pop();
    ans += nowans;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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;
struct Element {
  int value, cost;
  inline bool operator<(const Element &x) const { return value < x.value; }
};
Element a[200005];
priority_queue<int> heap;
int n, pnt, now, tmp;
long long ans, nowans;
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i].value);
  }
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i].cost);
  }
  sort(a + 1, a + n + 1);
  while (pnt < n || heap.empty() == false) {
    tmp++;
    if (heap.empty() == true) {
      tmp = a[pnt + 1].value;
    }
    while (pnt < n && a[pnt + 1].value == tmp) {
      pnt++;
      nowans += a[pnt].cost;
      heap.push(a[pnt].cost);
    }
    nowans -= heap.top();
    heap.pop();
    ans += nowans;
  }
  printf("%lld\n", ans);
  return 0;
}

```