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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <complex>
#include <iostream>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

using std::cin;
using std::cout;
using std::endl;
using std::cerr;

using std::string;
using std::to_string;
using std::vector;
using std::set;
using std::queue;
using std::stack;
using std::priority_queue;

using std::min;
using std::max;
using std::sort;
using std::abs;

typedef long long int ll;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;

int main() {
    int n;
    cin >> n;

    vector<int> a(n);
    for(int i = 0; i < n; i++){
        cin >> a[i];
    }

    int ans = INF;
    for(int i = -100; i <= 100; i++){
        int diff = 0;
        for(int j = 0; j < n; j++){
            diff += (a[j] - i) * (a[j] - i);
        }
        ans = min(ans, diff);
    }

    cout << ans << endl;

    return 0;
}