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.
import heapq
from collections import defaultdict


def solve(N, A, T):
    # Find conflicting nums
    numToConflicts = defaultdict(list)
    for i, x in enumerate(A):
        numToConflicts[x].append(i)

    # Iterate the conflict nums in heap since they are in a large range
    # Pick the best item to keep for each num, all other conflicts get scooted forward in another heap
    conflictNums = list(numToConflicts.keys())
    heapq.heapify(conflictNums)
    scoot = []
    assigned = 0
    cost = 0
    while assigned != N:
        num = heapq.heappop(conflictNums)

        heapCost = float("-inf")
        if scoot:
            heapCost, fromNum = scoot[0]
            heapCost *= -1

        conflictCost = float("-inf")
        indices = []
        if num in numToConflicts:
            indices = numToConflicts[num]
            maxIndex = max(indices, key=lambda i: T[i])
            conflictCost = T[maxIndex]

        # Pick costliest from conflicts or heap
        if heapCost > conflictCost:
            # Use the heap for this cell
            heapq.heappop(scoot)
            cost += (num - fromNum) * heapCost
        else:
            # Use the costliest of the existing indices for this cell
            indices = filter(lambda x: x != maxIndex, indices)
        assigned += 1

        # Scoot everything else into the heap (excluding max if that's the one that should be in this cell)
        for i in indices:
            heapq.heappush(scoot, (-T[i], num))
        # If there's something being scooted, make sure the next num is num + 1
        if scoot and (not conflictNums or conflictNums[0] != num + 1):
            heapq.heappush(conflictNums, num + 1)
    return cost


if __name__ == "__main__":
    N, = map(int, input().split())
    A = [int(x) for x in input().split()]
    T = [int(x) for x in input().split()]
    ans = solve(N, A, T)
    print(ans)
