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=[int(i) for i in input().split()]
T=[int(i) for i in input().split()]
AT=[A[i]*(10**6)+T[i] for i in range(n)]
AT.sort()
aa,tt=0,0
ans=0
H=[]
for at in AT:
  a,t=at//(10**6),at%(10**6)
  if a==aa:
    heapq.heappush(H,-t)
    tt+=t
  else:
    while H and aa<a:
      tm=heapq.heappop(H)
      tt+=tm
      ans+=tt
      aa+=1
    heapq.heappush(H,-t)
    aa=a
    tt+=t
while H:
  tm=heapq.heappop(H)
  tt+=tm
  ans+=tt
print(ans)