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 sys
from heapq import *

sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
f_inf = float('inf')
mod = 10 ** 9 + 7


def resolve():
    def val(l, nums):
        unit = l // nums
        extra = l - unit * nums
        return (nums - extra) * pow(unit, 2) + extra * pow(unit + 1, 2)

    n, k = map(int, input().split())
    A = list(map(int, input().split()))

    pq = []
    res = 0
    for i in range(n):
        res += pow(A[i], 2)
        heappush(pq, (-val(A[i], 1) + val(A[i], 2), A[i], 2))

    for _ in range(k - n):
        temp = heappop(pq)
        res += temp[0]
        a, b = temp[1], temp[2]
        heappush(pq, (-val(a, b) + val(a, b + 1), a, b + 1))

    print(res)


if __name__ == '__main__':
    resolve()
