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
from collections import deque
mod = 10**9 + 7
import sys
import random
# sys.setrecursionlimit(10**6)
from queue import PriorityQueue
# def rl():
#     return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
from itertools import permutations
from math import factorial as f

def ncr(x, y):
    return f(x) // (f(y) * f(x - y))

from collections import defaultdict
from heapq import heappop, heappush, heapify

n = int(input())
a = list(map(int, input().split()))
t = list(map(int, input().split()))
d = defaultdict(list)
for i in range(n):
    d[a[i]].append(t[i])
heap2 = list(set(a))
heapify(heap2)

ans = 0
used = set()
heap = []
s = 0
while heap2:
    cur = heappop(heap2)

    if cur in used:
        continue
    used.add(cur)

    for ti in d[cur]:
        heappush(heap, -ti)
        s += ti

    ti = -heappop(heap)
    # print(s,ti,cur)
    s -= ti
    ans += s

    if heap:
        heappush(heap2, cur + 1)
print(ans)
