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.
from heapq import *
n = int(input())
a = list(map(int,input().split()))
t = list(map(int,input().split()))
ls = list(zip(a,t))
ls.sort()
ls.append((10**18,0))
ans = 0
sm = 0
hq = []
xp = ls[0][0]
heapify(hq)
i = 0
while i<=n:
  if ls[i][0] == xp:
    heappush(hq,-ls[i][1])
    sm += ls[i][1]
    i += 1
  else:
    while hq:
      cnt = heappop(hq)
      cnt = -cnt
      ans += sm-cnt
      sm -= cnt
      xp += 1
      while i<=n:
        if ls[i][0] == xp:
          heappush(hq,-ls[i][1])
          sm += ls[i][1]
        elif ls[i][0] > xp:
          break
        i += 1
    xp = ls[i][0]
print(ans)