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 java.io.*;
import java.util.*;

public class Recommendations {
	public static void main(String[] args) throws NumberFormatException, IOException
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int N = Integer.parseInt(br.readLine());
		int[][] categories = new int[N][2];
		String[] temp = br.readLine().split(" ");
		for (int i = 0; i < N; i++)
			categories[i][0] = Integer.parseInt(temp[i]); // 0 is number of articles
		temp = br.readLine().split(" ");
		for (int i = 0; i < N; i++)
			categories[i][1] = Integer.parseInt(temp[i]); // 1 is cost
		Comparator<int[]> c = new Comparator<int[]>()
		{
			public int compare(int[] o1, int[] o2)
			{
				return o1[0]-o2[0];
			}
		};
		Arrays.sort(categories, c);
		Comparator<Integer> c2 = new Comparator<Integer>()
		{
			public int compare(Integer o1, Integer o2)
			{
				return o2-o1;
			}
		};		
		PriorityQueue<Integer> pq = new PriorityQueue<Integer>(c2);
		int current = categories[0][0]; long sum = 0; long ans = 0; int i = 0;
		while (i < N)
		{
			while (i < N && categories[i][0] == current)
			{
				pq.add(categories[i][1]);
				sum += categories[i][1]; i++;
			}
			if (!pq.isEmpty())
			{
				sum -= pq.remove();
				current++;
				ans += sum;
			}
			else
				current = categories[i][0];
			//System.out.println(pq);
			//System.out.println(sum);
		}
		while (!pq.isEmpty())
		{
			sum -= pq.remove();
			ans += sum;		
		}
		System.out.println(ans);
	}
}
