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.io.*;
  public  class Solution {
     static ArrayList<Integer> graph[];
     static int parent[];
     static boolean vis[];
     static int seive[]=new int[1000001];
     static int dis[]=new int[1000001];
     static int subtree[]=new int[1000001];
     static int N =1000001;
     static HashSet<String> hm=new HashSet<>();
   static HashSet<String> hm1=new HashSet<>();
     static int time=0;
     public static void main(String[] args) {
       Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
StringBuilder sb=new StringBuilder();
int n=sc.nextInt();
long a[]=new long[n];
long t[]=new long[n];
PriorityQueue<Pair> pq=new PriorityQueue<>(new myComparator());
for( int i=0;i<n;i++)
a[i]=sc.nextLong();

for( int i=0;i<n;i++)
t[i]=sc.nextLong();

Pair arr[]=new Pair[n];
for( int i=0;i<n;i++)
arr[i]=new Pair(a[i],t[i]);
Arrays.sort(arr, new Comparator<Pair>(){
public int compare(Pair a, Pair b)
{
return (int)(a.x-b.x);
}
});

long val=arr[0].x;
pq.add(arr[0]);
long sum=0;
sum=sum+arr[0].y;
long ans=0;
for( int i=0;i<n;)
{

if(pq.isEmpty())
{
i++;
if(i==n)
break;
pq.add(arr[i]);
sum+=arr[i].y;
val=arr[i].x;

}
while(i<n-1&&arr[i+1].x==val)
{
pq.add(arr[i+1]);
sum+=arr[i+1].y;
i++;
}
Pair d=pq.remove();
sum-=d.y;
ans+=sum;
val++;
if(i==n)
break;
}
sb.append(ans);
System.out.println(ans);
}
}
class Pair
{
long x,y;
Pair(long x, long y)
{
this.x=x;
this.y=y;
}
}
class myComparator implements Comparator<Pair>
{
public int compare(Pair a, Pair b)
{
return Long.compare(b.y, a.y);
}
}
