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.
from sys import stdin
from heapq import heappush, heappop
def f(i, j):
    p, q = divmod(i, j)
    return p * p * j + (2 * p + 1) * q
def main():
    n, k = map(int, stdin.readline().split())
    a = map(int, stdin.readline().split())
    q = []
    c = [1] * n
    ans = 0
    for i, x in enumerate(a):
        ans += x * x
        if x > 1:
            d = f(x, 2) - x * x
            heappush(q, (d, i))
    for _ in xrange(k - n):
        d, i = heappop(q)
        ans += d
        c[i] += 1
        x, t = a[i], c[i]
        if x > t:
            d = f(x, t + 1) - f(x, t)
            heappush(q, (d, i))
    print ans
main()
