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.
import sys
input = sys.stdin.readline
import heapq

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

score=n
ANS=0
for a in A:
    ANS+=a**2

def calc(x,sc):
    if sc==x:
        return x
    if sc>x:
        return 1<<60
    r,q=divmod(x,sc)
    return (sc-q)*(r**2)+q*((r+1)**2)

X=[]
for a in A:
    now=a**2
    net=calc(a,2)
    X.append((net-now,a,2))

heapq.heapify(X)

for i in range(k-score):
    #print(ANS,X)
    sa,a,sc=heapq.heappop(X)
    ANS+=sa
    now=calc(a,sc)
    net=calc(a,sc+1)

    heapq.heappush(X,(net-now,a,sc+1))

print(ANS)
    
