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.
a=int(input())
z=list(map(int,input().split()))
r=list(map(int,input().split()))
ans=[]
for i in range(a):
    ans.append([z[i],r[i]])
ans.sort()
from heapq import *
i=0
pq=[]
ss=0
sol=0
while(i<len(z)):

    if(len(pq)==0):
        heappush(pq,-1*ans[i][1])
        ss+=ans[i][1]
        v=ans[i][0]
        i+=1
    while(i<len(ans) and ans[i][0]==v):
        heappush(pq,-ans[i][1])
        ss+=ans[i][1]
        i+=1
    ss+=heappop(pq)
    sol+=ss
    v+=1
while(pq):
    ss+=heappop(pq)
    sol+=ss
print(sol)
        
    
