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.
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
 
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
 
def li():return [int(i) for i in input().rstrip('\n').split(' ')]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]

n = val()
l1 = li()
l2 = li()

d = defaultdict(list)
for i in range(n):
    d[l1[i]].append(l2[i])
# for i in d:d[i] = sorted(d[i])
cnt = Counter()
curr = l1[0]

l = []
for i in range(n):
    l.append([l1[i],l2[i]])
l.sort(key=lambda x: (x[0],-x[1]))

he = []
i = j = tot = 0
helpd = defaultdict(list)
# print(l,d)
while i < n:
    tru = 0
    if len(d[l[i][0]]):
        temp = l[i][0]
        while i < n and l[i][0] == temp:
            helpd[-l[i][1]].append(l[i][0])
            heapq.heappush(he,-l[i][1])
            d[l[i][0]].pop()
            i += 1

    else:
        tru = 1
    j = l[i-1][0] if i < n else l[-1][0]
    while len(he):
        if len(d[j]):break
        else:
            temp = heapq.heappop(he)
            tot += -(j - helpd[temp].pop())*temp
        j += 1
    if tru:i += 1
    # print(he,i,j,tot)
print(tot)
