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;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep2(i,n) for(int i=1;i<=(int)(n);i++)
#define rep3(i,n) for(int i=0;i<=(int)(n);i++)
#define REP(i,m,n) for(int i=m;i<(int)(n);i++)
#define repll(i,n) for(ll i=0;i<(ll)(n);i++)
#define repll2(i,n) for(ll i=1;i<=(ll)(n);i++)
#define REPLL(i,m,n) for(ll i=m;i<(ll)(n);i++)
#define INF (int)1e9  //
#define LINF (ll)4e18  //
#define MOD (int)(1e9)+7
#define MP make_pair
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define PI 3.1415926535898

//------------------------------------------------------

int main() {
    int n;
    cin >> n;
    int a[n];
    rep(i, n) cin >> a[i];

    int Min = INF;

    for (int i = -100; i <= 100; i++) {
        int sum = 0;

        rep(k, n) {
            sum += (a[k] - i)*(a[k] - i);
        }
        Min = min(Min, sum);
    }

    cout << Min << endl;
    return 0;
}