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 sys
import math
def i():return int(sys.stdin.readline().replace("\n",""))
def i2():return map(int,sys.stdin.readline().replace("\n","").split())
def s():return str(sys.stdin.readline().replace("\n",""))
def l():return list(sys.stdin.readline().replace("\n",""))
def intl():return [int(k) for k in sys.stdin.readline().replace("\n","").split()]
def lx():return list(map(lambda x:int(x)*-1,sys.stdin.readline().replace("\n","").split()))
def t():return tuple(map(int,sys.stdin.readline().replace("\n","").split()))

if __name__ == "__main__":pass

n = i()
a = intl()
if n == 1:print(0)
else:
    s = max(a)+min(a)
    k1 = s//2
    k2 = s//2 + s%2
    c1 = 0
    c2 = 0
    for i in range(n):
        c2 += (a[i]-k2)**2
        c1 += (a[i]-k1)**2
    print(min(c1,c2))