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 sys
input = sys.stdin.readline
import heapq

n = int(input())
x = list(map(int, input().split()))
t = list(map(int, input().split()))
a = []
M = 1000000
for i in range(n):
    a.append(x[i] * M + t[i])
a.sort()
i = 0
ans = 0
while i < n:
    xi, ti = divmod(a[i], M)
    j = i + 1
    while j < n:
        xj, tj = divmod(a[j], M)
        b = j - i + xi
        if xj >= b:
            break
        j += 1
    j -= 1
    if i < j:
        b = []
        ptr = i
        for target in range(xi, j - i + xi + 1):
            while ptr <= j:
                xp, tp = divmod(a[ptr], M)
                if xp <= target:
                    heapq.heappush(b, (-tp, xp))
                else:
                    break
                ptr += 1
            tp, xp = heapq.heappop(b)
            ans += -tp * (target - xp)
    i = j + 1

print(ans)