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 as hp

N = int(input())
A = list(map(int, input().split()))
T = list(map(int, input().split()))
AT = {}
for a, t in zip(A, T):
    if a in AT:
        AT[a].append(t)
    else:
        AT[a] = [t]

As = sorted(list(set(A)))

q = []
S = 0
ans = 0
ind = -1
for a in As:
    while q and ind <= a:
        s = - hp.heappop(q)
        S -= s
        ans += S
        ind += 1
    
    for t in AT[a]:
        hp.heappush(q, -t)
        S += t

    #s = - hp.heappop(q)
    #S -= s

    ind = a+1

while q:
    s = - hp.heappop(q)
    S -= s
    ans += S

print(ans)