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.
# Код программы написал на языке Python 3
import sys
from heapq import heappush, heappop

def main():
    n = int(sys.stdin.readline())
    h = sorted(list(zip([int(i) for i in sys.stdin.readline().split()], [int(i) for i in sys.stdin.readline().split()])))
    
    z, w, o, res = [], 0, 0, 0
    
    while o < n:
        
        t = h[o][0]
        w += h[o][1]
        heappush(z, -1 * h[o][1])
        
        while 0 < n - o - 1 and h[o][0] == h[o + 1][0]:
            o += 1
            w += h[o][1]
            heappush(z, -1 * h[o][1])
            
        if o + 1 == n:
            cur = 1e18
        else:
            cur = h[o + 1][0]
            
        while z and t - cur < 0:
            t += 1
            w += heappop(z)
            res += w
            
        o += 1
        
    print(res)
    
for test in range(1):
    main()