There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.

Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.

Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.

Input

The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.

The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.

It is guaranteed that the sum of a_i is at least k.

Output

Output one integer: the minimum sum of time taken for rabbits to eat carrots.

Examples

Input


3 6
5 3 1


Output


15


Input


1 4
19


Output


91

Note

For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15

For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
import java.util.*;
 
public class square{
  
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
      
    int n=sc.nextInt(),k=sc.nextInt();
   
    int[] size=new int[n];
    for(int i=0;i<n;i++)
    {size[i]=sc.nextInt();
    }
 int[] cuts=new int[n];PriorityQueue<Long> q=new PriorityQueue();long ans=0;
  for(int i=0;i<n;i++)
  {
cuts[i]=1;
ans+=cost(size[i],1);
long diff=cost(size[i],1)-cost(size[i],2);
q.add(-(diff*n+i));
  }
  int rab=n;
  while(rab<k)
  {
      long next=-q.poll();
      int ind=(int)(next%n);

ans-=next/n;
rab++;
cuts[ind]++;
long diff=cost(size[ind],cuts[ind])-cost(size[ind],cuts[ind]+1);
q.add(-(diff*n+ind));
  }
   System.out.println(ans);
    }
    static long cost(long size, long count) {
		long base = size/count;
		long xtr = size%count;
		return (count-xtr)*base*base+xtr*(base+1)*(base+1);
}
}