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.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    static int mod=(int)1e9+7;
    public static void main(String[] args) throws Exception {
        FastReader sc = new FastReader();
        int n=sc.nextInt();
        Node[] node=new Node[n];
        for(int i=0;i<n;i++){
            node[i]=new Node();
            node[i].a=sc.nextInt();
        }
        for(int i=0;i<n;i++)node[i].t=sc.nextInt();
        Arrays.sort(node, new Comparator<Node>() {
            @Override
            public int compare(Node o1, Node o2) {
                return o1.a-o2.a;
            }
        });
        PriorityQueue<Node> pq=new PriorityQueue<>(new Comparator<Node>() {
            @Override
            public int compare(Node o1, Node o2) {
                return o2.t-o1.t;
            }
        });
        long curr=node[0].a;
        long ans=0;
        for(int i=0;i<n;i++){
            if(curr==node[i].a){
                pq.add(node[i]);
                continue;
            }
            while(!pq.isEmpty() && curr<node[i].a){
                Node e=pq.poll();
                ans+=(curr-e.a)*e.t;
                curr++;
            }
            curr=node[i].a;
            pq.add(node[i]);
        }

        while(!pq.isEmpty()){
            Node e=pq.poll();
            ans+=(curr-e.a)*e.t;
      
            curr++;
        }
        System.out.println(ans);
    }
}

class Node{
    int a, t;
}
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;
    }
}
