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;

typedef long long int ll;

#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define all(a) a.begin(), a.end()
#define sz(a) (ll)(a.size())
#define endl "\n"

const ll L = 105;

vector <ll> a(L);

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);

    ll N;
    cin >> N;

    for(ll i=0; i<N; i++)
    {
        cin >> a[i];
    }

    ll cur = -100;
    ll cost = 1e15;

    while(cur<=100)
    {
        ll temp = 0;
        for(ll i=0; i<N; i++)
        {
            temp+=((cur-a[i])*(cur-a[i]));
        }

        cost = min(cost, temp);

        cur++;
    }

    cout << cost << endl;

    

    return 0;
}