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 TestClass {

	static PrintWriter out = new PrintWriter(System.out);
	public static void main(String[] args) throws IOException{
 
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(in.readLine());
		String s1[] = in.readLine().split(" ");
		String s2[] = in.readLine().split(" ");
		int arr[][] = new int[n][2];
		for(int i=0;i<n;i++) {
			int a = Integer.parseInt(s1[i]);
			int b = Integer.parseInt(s2[i]);		
			arr[i] = new int[] {a,b};
		}
		Arrays.sort(arr, (x,y)->Integer.compare(x[0], y[0]));
		PriorityQueue<Integer> pq = new PriorityQueue<Integer>((x,y)->Integer.compare(y, x));
		int i=0;
		long curr_sum=0;
		long ans=0;
		int curr = 0;
		while(i<n || !pq.isEmpty()){
			if(!pq.isEmpty()) {
				curr_sum -= pq.poll();
				ans += curr_sum;
				curr++;
			}
			else {
				curr = arr[i][0];
			}
			while(i<n && curr == arr[i][0]) {
				curr_sum += arr[i][1];
				pq.add(arr[i][1]);
				i++;
			}
		}
		out.println(ans);
        out.close();
	}
	
}