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
from collections import *
from heapq import *

n = int(input())
a = list(map(int, input().split()))
t = list(map(int, input().split()))
d = defaultdict(list)

for i in range(n):
    d[a[i]].append(t[i])

ks = list(d.keys())
ks.sort()
prev = ks[0]
ans = 0
s = 0
pq = []

for a in ks:
    for _ in range(a-prev-1):
        if len(pq)==0:
            break
        
        ans += s
        s -= -heappop(pq)
        
    ans += s
        
    for t in d[a]:
        heappush(pq, -t)
        s += t
    
    s -= -heappop(pq)
    prev = a
    
while len(pq):
    ans += s
    s -= -heappop(pq)

print(ans)