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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;


void solve(){

}

int main(){
    long long N;
    cin >> N;
    vector<long long>v(N);
    long long m = 0;
    for(int i = 0; i < N; i++){
        cin >> v.at(i);
        m += v.at(i);
    }
    m = m / N;

    long long m0 = m;
    long long m1 = m-1;
    long long m2 = m+1;
    long long s0 = 0;
    long long s1 = 0;
    long long s2 = 0;
    for(auto x: v){
        s0 += (x - m0)*(x-m0);
        s1 += (x - m1)*(x-m1);
        s2 += (x - m2)*(x-m2);
    }
    cout << min(s0, min(s1,s2)) << endl;
    

}