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.io.*; import java.util.*;
public class CF1428E {
    //Another problem where we analyze the gains of each move
    //Similar to
    //USACO Plat (Building a tall barn)
    public static class Pair implements Comparable<Pair>{
        long val; int ind;
        public Pair(long a, int b){
            this.val=a; this.ind=b;
        }
        public int compareTo(Pair other){
            if(this.val>other.val)return -1;
            if(this.val<other.val)return 1;
            if(this.ind>other.ind)return 1;
            return -1;
            
        }
    }
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        int N=Integer.parseInt(st.nextToken());
        int K=Integer.parseInt(st.nextToken());
        int[] a=new int[N];
        st=new StringTokenizer(br.readLine()); 
        long ans=0;
        for (int i = 0; i < N; i++) {
            a[i]=Integer.parseInt(st.nextToken()); 
            ans+=(long) a[i]*a[i];
        }
        int[] sp=new int[N]; Arrays.fill(sp,1);
        //Note repeatedly splitting, by convexity, gives it less and less reduction
        PriorityQueue<Pair> pq=new PriorityQueue<>();
        for (int i = 0; i < N; i++) {
            pq.add(new Pair(f(a[i],1)-f(a[i],2),i));//Make sure they're negative
        }
        int t=K-N;
        while(t>0){
            t--;
            Pair p=pq.poll();
            ans-=p.val; sp[p.ind]++;
            pq.add(new Pair(f(a[p.ind],sp[p.ind])-f(a[p.ind],sp[p.ind]+1),p.ind));
        }
        System.out.println(ans);
    }
    public static long f(int k, int s){
        long q=k/s; long r=k%s;
        return (long) (s-r)*q*q+r*(q+1)*(q+1);
    }
}
