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())
 
l1 = [int(i) for i in input().split()]
l2 = [int(i) for i in input().split()]
l = list(zip(l1, l2))
l.sort(key=lambda x: (x[0], -x[1]))

 
h = []
curr = l[0][0]
i = 1
res = 0
while (i < n):
 
    while (h):
        if (curr >= l[i][0] - 1):
            break
        curr += 1
        temp = heapq.heappop(h)
        res += (curr - temp[1]) * (-temp[0])
 
    heapq.heappush(h, [-l[i][1], l[i][0]])
 
    if (curr < l[i][0]):
        curr = l[i][0]
        temp = heapq.heappop(h)
        res += (curr - temp[1]) * (-temp[0])
    i += 1
 
 
while (h):
    curr += 1
    temp = heapq.heappop(h)
    res += (curr - temp[1]) * (-temp[0])
 
print(res)