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


def f(length, count):
    u, v = divmod(length, count)
    return (count-v)*u**2 + v*(u+1)**2


def main():

    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    res = sum(v**2 for v in a)
    b = [(-f(v, 1)+f(v, 2), v, 1) for v in a]
    heapq.heapify(b)
    for i in range(k-n):
        dif, v, count = heapq.heappop(b)
        res += dif
        heapq.heappush(b, (-f(v, count+1)+f(v, count+2), v, count+1))
    print(res)
    return


if __name__ == "__main__":
    main()

