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)

#define _USE_MAT_DEFINES
#include<iostream>
#include <iomanip>
#include<stdio.h>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
#include <queue>
#include <utility>
#include <map>
#include <set>
#include <deque>
#include <numeric>
#include <functional>
using namespace std;

long long cost(int x, int y) {
    return (x - y)*(x - y);
}

int main() {
    int N, min_ = 100, max_ = -100;
    cin >> N;
    vector<int>a(N);
    for (int i = 0; i < N; i++) {
        cin >> a[i];
        min_ = min(min_, a[i]);
        max_ = max(max_, a[i]);
    }

    long long ans = 1e9;
    for (int i = min_; i <= max_; i++) {
        long long c = 0;
        for (int j = 0; j < N; j++) {
            c += cost(a[j], i);
        }
        ans = min(ans, c);
    }

    cout << ans << endl;
    
    return 0;
}