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 Julia (0.5.0) (judge verdict: Accepted)

MyString=Union{String,SubString{String},Char}
MyStringArray=Union{Array{String,1},Array{SubString{String},1},Array{Char,1}}
parseInt(x::MyString) = parse(Int, x)
parseInt(x::Real) = trunc(Int, x)
parseBigInt(x::MyString) = parse(BigInt, x)
parseBigInt(x::Real) = trunc(BigInt, x)
parseFloat(x::MyString) = parse(Float64, x)
parseFloat(x::Real) = trunc(Float64, x)
parseBigFloat(x::MyString) = parse(BigFloat, x)
parseBigFloat(x::Real) = trunc(BigFloat, x)
parseMap(x::MyStringArray) = map(parseInt, x)
parseMapBig(x::MyStringArray) = map(parseBigInt, x)

function main()
    n=parseInt(readline())
    a=sort(parseMap(split(readline())))
    ans=10^9
    for i in a[1]:a[end]
        res=0
        for j in a
            res+=(i-j)^2
        end
        ans=min(res,ans)
    end
    println(ans)
end
main()