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 sys
input = sys.stdin.readline
n = int(input())
raw1 = list(map(int,input().split()))
raw2 = list(map(int,input().split()))
c = []
for i in range(n):
  c.append((raw1[i],raw2[i]))
c.sort()
c.append((10**12,0))
#print(c)
import heapq
ans = 0
last = 0
now = []
for x in c:
  a,t = x
  for ai in range(last,a):
    if not now:
      break
    tm,am = heapq.heappop(now)
    tm = 10**5-tm
    ans += (ai-am) * tm
    #print(ai-am,tm)
  heapq.heappush(now,(10**5-t,a))
  last = a
  #print(ans,now)
#print(ans,now)
print(ans)