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

public class Abc {
    public static void main(String[] args) throws Exception {
        FastReader sc = new FastReader();
        int n=sc.nextInt();
        Point p[]=new Point[n];
        for (int i=0;i<n;i++)p[i]=new Point(sc.nextInt(),0);
        for (int i=0;i<n;i++)p[i].y=sc.nextInt();
        Arrays.sort(p, new Comparator<Point>() {
            @Override
            public int compare(Point o1, Point o2) {
                return o1.x-o2.x;
            }
        });
        PriorityQueue<Integer> pq=new PriorityQueue<>(Collections.reverseOrder());
        int cur=-1;
        int pos=0;
        long sum=0;
        long ans=0;
        while (!pq.isEmpty() || pos<n){
            if (pq.isEmpty())cur=p[pos].x;
            while (pos<n && cur==p[pos].x){
                pq.add(p[pos].y);
                sum+=p[pos].y;
                pos++;
            }
            sum-=pq.poll();
            ans+=sum;
            cur++;

        }
        System.out.println(ans);
    }
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}