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
import sys
import heapq

q = []

n,k = map(int,stdin.readline().split())
a = list(map(int,stdin.readline().split()))
x = [1] * n

for ind in range(n):
    s = a[ind] // x[ind]
    rem = a[ind] % x[ind]
    score = ((s+1)**2)*rem + (s**2)*(x[ind]-rem)

    s2 = a[ind] // (x[ind]+1)
    rem2 = a[ind] % (x[ind]+1)
    score2 = ((s2+1)**2)*rem2 + (s2**2)*((x[ind]+1)-rem2)

    q.append( (score2-score ,ind ) )

heapq.heapify(q)

for loop in range(k-n):
    
    tmp,ind = heapq.heappop(q)
    x[ind] += 1

    s = a[ind] // x[ind]
    rem = a[ind] % x[ind]
    score = ((s+1)**2)*rem + (s**2)*(x[ind]-rem)

    s2 = a[ind] // (x[ind]+1)
    rem2 = a[ind] % (x[ind]+1)
    score2 = ((s2+1)**2)*rem2 + (s2**2)*((x[ind]+1)-rem2)
    
    heapq.heappush( q , (score2-score , ind) )

ans = 0
#print (x)
for i in range(n):
    s = a[i] // x[i]
    rem = a[i] % x[i]

    #print (ans,((s+1)**2)*rem , (s**2)*(x[i]-rem))

    ans += ((s+1)**2)*rem + (s**2)*(x[i]-rem)
print (ans)