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.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
  
    void solve() throws IOException {
        int n=ni();
        int[][]A=new int[n+1][2];
        for (int x=0;x<n;x++) A[x][0]=ni();
        for (int x=0;x<n;x++) A[x][1]=ni();
        A[n][0]=Integer.MAX_VALUE;
        Arrays.sort(A,(a,b)->Integer.compare(a[0],b[0]));
        
        long ans=0;
        long cost=A[0][1];
        int p=0;
        int q=1;
        int curr=A[0][0];
        PriorityQueue<Integer>PQ=new PriorityQueue(Collections.reverseOrder());
        PQ.add(A[0][1]);
        
        while (true) {
            while (q<n && A[q][0]==curr) { PQ.add(A[q][1]); cost+=A[q][1]; q++; }
            p++;
            if (p==q) {
                q++;
                if (q>=n) break;
                PQ.clear();
                cost=A[p][1];
                PQ.add(A[p][1]);
                curr=A[p][0];
                continue;
            }
            cost-=PQ.poll();
            //System.out.println(cost);
            ans+=cost;
            curr++;
        }
        
        System.out.println(ans);
    }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}