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.
def calc(l, n):
    smol = l // n
    tol = l % n
    return smol * smol * n + (2 * smol + 1) * tol


import heapq
n, k = map(int, input().split())
a = list(map(int, input().split()))
curr = [1] * n

q = []
for i in range(n):
    v = a[i]
    heapq.heappush(q, (calc(v, 2) - calc(v, 1), i))

out2 = sum(map(lambda x: x*x, a))

for i in range(k - n):
    diff, nex = heapq.heappop(q)
    curr[nex] += 1
    v = a[nex]
    heapq.heappush(q, (calc(v, curr[nex] + 1) - calc(v, curr[nex]), nex))
    out2 += diff

out = 0
for i in range(n):
    out += calc(a[i], curr[i])
assert(out == out2)
print(out)
