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 heapq import heapify, heappop, heappush


def getCost(num, k):
	part = num // k
	big = num % k
	small = k - big
	return small * part * part + big * (part + 1) * (part + 1)


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

ar = []
ans = 0
for i in range(n):
	ans += ls[i] * ls[i]
	diff = getCost(ls[i], 1) - getCost(ls[i], 2)
	ar.append([-diff, ls[i], 1])
heapify(ar)
i = 0
while i < k - n:
	elem = heappop(ar)
	ans -= (-elem[0])
	nxt = [-1*(getCost(elem[1], elem[2] + 1) - getCost(elem[1] ,elem[2] + 2)), elem[1], elem[2] + 1]
	heappush(ar, nxt)
	i += 1

print(ans)
