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.*;

/**
 *
 * @author True
 */

public class Main {
    public static Long calc(int n, int k) {
        Long s = Long.valueOf(n/k);
        Long t = Long.valueOf(n%k);
        return t*(s+1)*(s+1) + (k-t)*s*s;
    }
    public static int[] a = new int[100010];
    public static int[] b = new int[100010];
    public static void main(String args[]) {
        Scanner in;
        PrintStream out;
        in = new Scanner (new BufferedInputStream(System.in));
        out = System.out;
        try{
            in = new Scanner(new FileInputStream("java.in"));
            out = new PrintStream("java.out");
        } catch (FileNotFoundException e) {}
        TreeSet<Long> st;
        st = new TreeSet<>();
        int n = in.nextInt(), k = in.nextInt();
        Long tmp = (long)1000000;
        Long ans = (long)0;
        for(int i=1; i<=n; i++) {
            b[i] = 1;
            a[i] = in.nextInt();
            Long f = calc(a[i], 2) - calc(a[i], 1);
            f *= tmp;
            f += i;
            st.add(f);
            ans += (long)a[i]*a[i];
        }
        for(int i=0; i<k-n; i++) {
            Long f = st.first();
            int g = (int)(f%tmp);
            if(g<0) g+=tmp;
            ans += (f-g)/tmp;
            b[g]++;
            st.remove(f);
            f = calc(a[g], b[g]+1) - calc(a[g], b[g]);
            f *= tmp;
            f += g;
            st.add(f);
        }
        out.println(ans);
    }
}
 