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.util.*;
import java.math.*;

public class HelloWorld {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int num = scanner.nextInt();
        int[] a = new int[num];
        int[] t = new int[num];
        int[][] pt = new int[num][2];
        for(int i = 0; i < num; i++) {
            a[i] = scanner.nextInt();
        }
        for(int i = 0; i < num; i++) {
            t[i] = scanner.nextInt();
            pt[i] = new int[]{a[i], t[i]};
        }
        Arrays.sort(pt, (x, y) -> Integer.compare(x[0], y[0]));
        PriorityQueue<Integer> pq = new PriorityQueue<>((x, y) -> -Integer.compare(x, y));

        int pos = 0;
        long curr_sum = 0;
        long ans = 0;
        long curr = 0;

        while(!pq.isEmpty() || pos < num) {
            if(!pq.isEmpty()) {
                curr_sum -= pq.poll();
                ans += curr_sum;
                curr++;
            } else {
                curr = pt[pos][0];
            }

            while(pos < num && pt[pos][0] == curr) {
                pq.add(pt[pos][1]);
                curr_sum += pt[pos][1];
                pos++;
            }
        }
        System.out.println(ans);
    }
}