1310_A. Recommendations
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.
```python
import heapq
n = int(input())
*a, = map(int, input().split())
*t, = map(int, input().split())


i = 0

sorted_idxs = sorted(range(n), key=lambda o: (a[o], -t[o]))
cur_val = a[sorted_idxs[0]]
dupls = []
ans = 0
s = 0
for i, idx in enumerate(sorted_idxs):
    cur_val = a[idx]
    heapq.heappush(dupls, (-t[idx], a[idx]))
    s += t[idx]
    while dupls and (i == len(a) - 1 or cur_val + 1 <= a[sorted_idxs[i + 1]]):
        cur_val += 1
        largest, a_val = heapq.heappop(dupls)
        largest *= -1
        s-= largest
        ans += s

print(ans)

```