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 sys import stdin, gettrace
from heapq import *

if not gettrace():
    def input():
        return next(stdin)[:-1]


def main():
    n = int(input())
    aa = [int(a) for a in input().split()]
    bb = [int(a) for a in input().split()]
    bss = {}
    aheap = []

    def insert_bss(a, b):
        if a not in bss:
            bss[a] = []
            heappush(aheap, a)
        bss[a].append(b)

    for a, b in zip(aa, bb):
        insert_bss(a, b)
    cost = 0
    moving = []
    stepcost = 0
    while aheap:
        a = heappop(aheap)
        if len(bss[a]) > 1:
            bss[a].sort()
            for b in bss[a][:-1]:
                heappush(moving, -b)
                stepcost+=b
        cost += stepcost
        if len(moving) > 0:
            bi = -heappop(moving)
            insert_bss(a+1, bi)
            stepcost -= bi
    print(cost)



if __name__ == "__main__":
    main()