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)

//#pragma GCC optimize(\"Ofast,no-stack-protector\")
//#pragma GCC target(\"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx\")
//#pragma GCC target(\"avx,tune=native\")
// Neil

#include<bits/stdc++.h>

using namespace std;

typedef complex<double> base;
typedef long double ld;
typedef long long ll;

#define endl "\n"
#define pb push_back
#define pii pair<int,int>
#define pll pair< ll , ll >
#define vi vector<int>
#define vvi vector< vi >
const int MAXN=(int)(2e5+5);
const ll MOD=(ll)(1e9+7);
int t;
int arr[MAXN];
int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);
    ll n;
    cin>>n;
    ll s = 0;
    for(int i=0;i<n;i++){
        cin>>arr[i];
        s+=arr[i];
    }
    ll a = (ll)s/n;
    
    if(s%n == 0){
        int cost = 0;
        for(int i=0;i<n;i++){
            cost += 1ll*(arr[i]-a) * (arr[i]-a);
        }
        cout<<cost<<endl;
    }
    else{
        int cost1 = 0, cost2 = 0;
        for(int i=0;i<n;i++){
            cost1 += 1ll*(arr[i]-a) * (arr[i]-a);
        }
        a++;
        for(int i=0;i<n;i++){
            cost2 += 1ll*(arr[i]-a) * (arr[i]-a);
        }
        cout<<min(cost1, cost2)<<endl;
    }
    return 0;
}