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.*;
import java.math.*;

public class E {
	public static void main(String[] args) throws IOException {

		/**/
		Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		/*/
		Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(new FileInputStream("src/e.in"))));
		/**/
		
		int n = sc.nextInt();
		int k = sc.nextInt();
		long[] shark = new long[n];
		for (int i = 0; i < n; ++i)
			shark[i] = sc.nextInt();
		long ans = 0;
		int[] div = new int[n];
		
		PriorityQueue<Long> pq = new PriorityQueue<>();
		for (int i = 0; i < n; ++i) {
			div[i] = 1;
			ans += calc(shark[i],1);
			long diff = calc(shark[i],1)-calc(shark[i],2);
			pq.add(-(diff*n+i));
		}
		int rab = n;
		while (rab < k) {
			long rem = -pq.remove();
			int ind = (int)(rem%n);
			long cd = rem/n;
			ans -= cd;
			++rab;
			++div[ind];
			long nd = calc(shark[ind],div[ind])-calc(shark[ind],div[ind]+1);
			pq.add(-(nd*n+ind));
		}
		System.out.println(ans);
	}
	
	public static long calc(long size, long count) {
		long base = size/count;
		long xtr = size%count;
		return count*base*base+xtr*(base*2+1);
	}
}