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 _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define ALL(v) v.begin(),v.end()
using ll = long long;
using pii = pair<int,int>;
using vi = vector<int>;
using vs = vector<string>;
using vll = vector<long long>;
using vpii = vector<pair<int,int>>;
using vivi = vector<vector<int>>;
const int dx[4] = {1,0,-1,0};
const int dy[4] = {0,1,0,-1};
const int mod = 10000007;
const int inf = 1001001001;



int main(void) {
  int n;
  cin >> n;
  vi a(n);
  rep(i,n) cin >> a[i];
  ll sum = inf;
  rep2(j,-100,101){
    ll kamo = 0;
    rep(i,n){
      kamo += (a[i]-j)*(a[i]-j);
    }
    sum = min(sum,kamo);
  }
  cout << sum << endl;
}