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 heapq
n=int(input())
a=list(map(int,input().split()))
t=list(map(int,input().split()))
d={}
for i in range(len(a)):
    if a[i] not in d:
        d[a[i]]=[]
    d[a[i]].append(t[i])
k=list(d.keys())
heapq.heapify(k)
cost=0
mxh=[]
sum_=0
while k:
    key=heapq.heappop(k)
    for i in d[key]:
        sum_+=i 
        heapq.heappush(mxh,-i)
    
    if not mxh:
        continue
    sum_+=heapq.heappop(mxh)
    cost+=sum_
    
    if not mxh:
        continue
    if key+1 not in d:
        heapq.heappush(k,key+1)
        d[key+1]=[]
    del d[key]
        
print(cost)
    