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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
using lli = int_fast64_t;
using ulli = uint_fast64_t;
#define debug(x) cerr << #x << " : " << x << endl
#define debugv(x) cerr << #x << " : {";for(auto value_of_x:x){ cerr << value_of_x << ", ";}cerr << endl
#define debugst(x) auto xt = x;cerr << #x << " : {";for(int i=0;i<x.size();i++){ cerr << xt.top() << ", ";xt.pop();}cerr << endl
#define debugvp(x) cerr << #x << " : {";for(auto value_of_x:x){ cerr << value_of_x.first << " : " << value_of_x.second << ", ";}cerr << endl
#define HYPP(x) hypot(x.first,x.second)

constexpr lli MOD = 1e9+7;
constexpr lli LIMIT = 1000000000000000000 * 5;
constexpr int tenfive = 100000;


int main() {
	int N;
	cin >> N;
	vector<int> v(N);
	for (int i = 0; i < N; i++) {
		cin >> v[i];
	}


	int ans = MOD;
	for (int i = 0; i < 100; i++) {
		int sum = 0;
		for (int j = 0; j < N; j++) {
			sum += pow(i - v[j], 2);
		}
		ans = min(ans, sum);
	}

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