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 *

def find_add(a, k):
    extra = a%k
    return (k-extra)*((a//k)**2) + extra*((1+(a//k))**2)

n, mx = map(int, input().split())
lst = list(map(int, input().split()))
a = []
heapify(a)
    
for i in lst:
    heappush(a, (find_add(i, 2)-find_add(i, 1), i, 1))
    
total = n

while total != mx:
    curr_cont, el, k = heappop(a)
    new = (find_add(el, k+2)-find_add(el, k+1), el, k+1)
    heappush(a, new)
    total += 1

sc = 0 
while a:
    curr_cont, el, k = heappop(a)
    sc += find_add(el, k)
    
print(sc)