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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class TaskD {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        PrintWriter pw = new PrintWriter(System.out);
        
        int n = Integer.parseInt(br.readLine());
        Pair[] p = new Pair[n];
        st = new StringTokenizer(br.readLine());
        for(int i=0 ; i<n ; i++) {
            int x = Integer.parseInt(st.nextToken());
            p[i] = new Pair(x, 0);
        }
        st = new StringTokenizer(br.readLine());
        for(int i=0 ; i<n ; i++) p[i].y = Integer.parseInt(st.nextToken());

        Arrays.sort(p, (o1, o2) -> o1.x-o2.x);

        PriorityQueue<Pair> pq = new PriorityQueue<>(10, (o1, o2) -> o2.y-o1.y);

        int start = p[0].x;

        long ans = 0;

        for(int i=0 ; i<n ; ) {
            int curr = p[i].x;
            start = p[i].x;
            pq.add(p[i]);
            i++;
            while(i<n && p[i].x == p[i-1].x) {
                pq.add(p[i]);
                i++;
            }
            if(i == n) {
                while(!pq.isEmpty()) {
                    Pair x = pq.poll();
                    ans += (start-x.x)*(long)x.y;
                    start++;
                }
            }
            else {
                while(start != p[i].x && !pq.isEmpty()) {
                    Pair x = pq.poll();
                    ans += (start-x.x)*(long)x.y;
                    start++;
                }
            }

        }

        pw.println(ans);
        
        
        
        
        
        pw.flush();
        pw.close();
    }
}

class Pair {
    int x;
    int y;
    Pair(int a, int b) {
        x = a;
        y = b;
    }
}
