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.

import collections
union_arr =collections.defaultdict(int)
def func():
    n_cates = int(input())
    cate_i = list(map(int,input().split()))
    time_i = list(map(int,input().split()))
    idx = [index for index,value in sorted(list(enumerate(time_i)),reverse=True,key=lambda x:x[1])]
    cate_i = [cate_i[i] for i in idx]
    time_i = [time_i[i] for i in idx]
    ans = 0
    for i in range(n_cates):
        f = find(cate_i[i])
        if f == cate_i[i]:
            union(f,f+1)
        else:
            union(f,f+1)
            ans +=(f-cate_i[i])*time_i[i]

    print(ans)

def union(a,b):
    rt_b = find(b)
    if rt_b!=a:
        union_arr[a]=rt_b

def find(a):
    if union_arr[a]==0:
        return a 
    union_arr[a] = find(union_arr[a])
    return union_arr[a]

    

if __name__ == "__main__":
    func()