Problem:
Score : 200 points

Problem Statement

Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.

He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).

Find the minimum total cost to achieve his objective.

Constraints

1≦N≦100

-100≦a_i≦100

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum total cost to achieve Evi's objective.

Sample Input 1

2
4 8

Sample Output 1

8

Transforming the both into 6s will cost (4-6)^2+(8-6)^2=8 dollars, which is the minimum.

Sample Input 2

3
1 1 3

Sample Output 2

3

Transforming the all into 2s will cost (1-2)^2+(1-2)^2+(3-2)^2=3 dollars. Note that Evi has to pay (1-2)^2 dollar separately for transforming each of the two 1s.

Sample Input 3

3
4 2 5

Sample Output 3

5

Leaving the 4 as it is and transforming the 2 and the 5 into 4s will achieve the total cost of (2-4)^2+(5-4)^2=5 dollars, which is the minimum.

Sample Input 4

4
-100 -100 -100 -100

Sample Output 4

0

Without transforming anything, Evi's objective is already achieved. Thus, the necessary cost is 0.

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

import numpy as np
N = int(input())
a = [int(i) for i in input().split()]
oppai = np.array(a)

max = np.max(oppai)
min = np.min(oppai)
x = (max+min)//2
sum = [0] * 3

for i in range(x-1,x+2,1):
    oppai = np.array(a) #oppaiの初期化
    oppai -= i
    oppai = oppai**2
    sum[i-x+1] = np.sum(oppai)
    #print('sum[{}]'.format(i-x+1),sum[i-x+1])

chinko = np.array(sum)
#print(np.argmin(chinko)+x-1)
print(np.min(chinko))