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 sys import stdin,stderr
def rl():
    return [int(w) for w in stdin.readline().split()]

n, = rl()
a = rl()
t = rl()

from collections import defaultdict
from heapq import *

cd = defaultdict(list)
for ti, ai in zip(t, a):
    heappush(cd[ai], -ti)
ch = []
for a, th in cd.items():
    heappush(ch, (a,-sum(th),th))
r = 0
while ch:
    a, ts, th = heappop(ch)
    mt = heappop(th)
    ts += mt
    r += ts
    if not th:
        continue
    a += 1
    if ch:
        na, nts, nth = heappop(ch)
        if na > a:
            heappush(ch, (na, nts, nth))
        else:
            ts += nts
            if len(th) < len(nth):
                th, nth = nth, th
            for mt in nth:
                heappush(th, mt)
    heappush(ch, (a, ts, th))
print(r)
