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.


"""
NTC here
""" 
import sys
inp= sys.stdin.readline
input = lambda : inp().strip()
# flush= sys.stdout.flush
# import threading
# sys.setrecursionlimit(10**6)
# threading.stack_size(2**26)

def iin(): return int(input())
def lin(): return list(map(int, input().split()))



def main():
    import heapq as hq 
    n = iin()
    a = lin()
    t = lin()
    dc = {}
    for i in range(n):
        try:
            dc[a[i]].append(t[i])
        except:
            dc[a[i]]=[t[i], ]
    # print(dc)
    sa = list(dc.keys())
    sa.sort()
    n = len(sa)
    ch = 0
    pt = sa[ch]
    temp = []
    sm = 0
    ans = 0
    hq.heapify(temp)
    done = 0
    while ch<n:
        # print(temp, sa[ch])
        pt = sa[ch]
        if len(dc[pt])>1 or done:
            a1 = sorted(dc[pt])
            for item in a1:
                hq.heappush(temp, -item)
                sm += item
            x = hq.heappop(temp)*(-1)
            dc[pt]=[x]
            sm -= x
            while len(temp):
                #print('A', ch, pt, temp )
                ans += sm
                pt += 1
                if pt in dc:
                    ch+=1
                    done = 1
                    break
                else:
                    x = hq.heappop(temp)*(-1)
                    sm -= x
            else:
                done = 0
        else:
            ch+=1
            done = 0
    # print(dc)
    print(ans)





main()

#threading.Thread(target=main).start()

