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 heapq
# import sys
# input = iter(sys.stdin.read().splitlines()).__next__

def time(l, p):
    """
    time taken to eat carrot of length l split into p pieces
    """
    rounded_up = l % p
    rounded_down = p - rounded_up
    q = l//p
    return rounded_down * q**2 + rounded_up * (q+1)**2

def f(l, p):
    """
    (nonpositive) difference between times for carrot of length l if split
    into p or p+1 pieces
    """
    return time(l, p+1) - time(l, p)

n, k = map(int, input().split())
a = [int(i) for i in input().split()]

splits = [1]*n
queue = [(f(a_i, splits[i]), i) for i, a_i in enumerate(a)]
heapq.heapify(queue)
total = sum(a_i*a_i for a_i in a)
for _ in range(k-n):
    f_i, i = heapq.heappop(queue)
    splits[i] += 1
    total += f_i
    heapq.heappush(queue, (f(a[i], splits[i]), i))
print(total)