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

# for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
ts = list(map(int, input().split()))

tt = [(ts[i], i) for i in range(n)]
tt.sort(reverse=True)

conflict = {}
next = {}
money = 0


def get_next(i):
	if i not in conflict:
		return i

	if next[i] in conflict:
		next[i] = get_next(next[i])

	return next[i]

for tx in tt:
	t, i = tx
	nee = get_next(a[i])
	money += (nee - a[i])  * t
	conflict[nee] = True
	next[nee] = nee+1

print(money)

