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 collections import deque
import heapq
import sys
input = lambda: sys.stdin.readline().strip()
n = int(input())
b = map(int,input().split())
c = map(int,input().split())
d = list(zip(b,c))
d.sort()
h = []
s = 0
i = 0
ans = 0
while i <n:
    lvl = d[i][0]
    s+=d[i][1]
    heapq.heappush(h, -d[i][1])
    while i+1<n and d[i][0]==d[i+1][0]:
        i+=1
        s+=d[i][1]
        heapq.heappush(h,-d[i][1])
    if i+1==n:
        lol = 10000000000000000000
    else:
        lol = d[i+1][0]
    while h and lvl<lol:
        lvl+=1
        x = heapq.heappop(h)
        s+=x
        ans+=s
    i+=1
print(ans)

# for i in range(n-1)
