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)

#pragma once
#include <sstream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
#include <iostream>
#include <utility>
#include <set>
#include <cctype>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <numeric>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 1000000007;
const int MAX_V = 201;
int now = 0;
int deco[1000];

int c[30];
ll dp[52][2510];
ll a[101];
void Solve() {
	int n;
	cin >> n;
	ll res = INF;
	for (size_t i = 0; i < n; i++)
	{
		cin >> a[i];
	}
	sort(a, a + n);
	

	for (size_t i = a[0]; i <= a[n - 1]; i++)
	{
		ll val = 0;
		for (size_t j = 0; j < n; j++)
		{
			val += (a[j] - i) * (a[j] - i);
		}
		res = min(res, val);
	}

	cout << (a[0] == a[n-1] ? 0 : res) << endl;
}
int main() {
	Solve();
	return 0;
}