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 collections import defaultdict
import heapq

n = int(input())
a = list(map(int, input().split()))
t = list(map(int, input().split()))

d = defaultdict(list)

for i in range(n):
    d[a[i]].append(i)

h_a = list(d.keys())
heapq.heapify(h_a)

h_t = []

time = 0
s = 0
done = set()

while h_a:
    cur_a = heapq.heappop(h_a)
    if cur_a in done:
        continue
    
    done.add(cur_a)

    for idx in d[cur_a]:
        heapq.heappush(h_t, -t[idx])
        s += t[idx]

    max_t = -heapq.heappop(h_t)
    s -= max_t

    time += s

    if h_t:
        heapq.heappush(h_a, cur_a+1)


print(time)



