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, bisect_right as br, insort
import sys
import heapq
from math import *
from collections import defaultdict as dd, deque
def data(): return sys.stdin.readline().strip()
def mdata(): return map(int, data().split())
#sys.setrecursionlimit(100000)

n=int(data())
A=list(mdata())
T=list(mdata())
AT=sorted(list(zip(A,T)),key=lambda x: (x[0],x[1]))
H=[]
a=ans=0
s=0
for i in AT:
    if i[0]==a:
       heapq.heappush(H,-i[1])
       s+=i[1]
    else:
        if H:
            t=heapq.heappop(H)
            s+=t
            a+=1
        while H and a!=i[0]:
            t=heapq.heappop(H)
            ans+=s
            s+=t
            a+=1
        ans+=s
        heapq.heappush(H,-i[1])
        s+=i[1]
        a=i[0]

if H :
    t=heapq.heappop(H)
    s+=t
while H :
    t=heapq.heappop(H)
    ans+=s
    s+=t

print(ans)

