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
import bisect as bi
import collections as cc
import heapq as hp
input=sys.stdin.readline
I=lambda:list(map(int,input().split()))
def ch(n,k):
    temp=n//k
    still=n%k
    ans=(k-still)*((temp)**2)+still*((temp+1)**2)
    return ans
n,k=I()
l=I()
ans=0
heap=[]
for i in range(n):
    ans+=l[i]**2
    hp.heappush(heap,(-ch(l[i],1)+ch(l[i],2),l[i],2))
for i in range(abs(n-k)):
    now=hp.heappop(heap)
    ans+=now[0]
    x,y=now[1],now[2]
    hp.heappush(heap,(-ch(x,y)+ch(x,y+1),x,y+1))
print(ans)