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.
n = int(input())
al = list(map(int,input().split()))
tl = list(map(int,input().split()))

atl = []
for a,t in zip(al, tl):
    atl.append((a, -t))
atl = sorted(atl)
kouho = []
import heapq
heapq.heapify(kouho)
ans = 0
s = 0
idx = 0
while True:
    if idx == n:
        while kouho:
            t1, a1 = heapq.heappop(kouho)
            ans += (s - a1) * t1
            s += 1
        print(-ans)
        exit()
    if atl[idx][0] <= s:
        a1, t1 = atl[idx]
        heapq.heappush(kouho, (t1, a1))
        idx += 1
        continue
    elif kouho:
        t1, a1 = heapq.heappop(kouho)
        ans += (s - a1) * t1
        s += 1
    else:
        a1, t1 = atl[idx]
        heapq.heappush(kouho, (t1, a1))
        idx += 1
        s = a1
        continue



