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)

// glory be to the sky
#include<cstdio>
#include<bits/stdc++.h> 
#define pb push_back
using namespace std;
	
typedef long long ll; 

int main(){

	ios::sync_with_stdio(false);
	cin.tie(0);

	int N, arr[105];
	ll ans = (ll) 1e9;
	cin >> N;
	for (int i = 0; i < N; i++) cin >> arr[i];

	for (int i = 0; i < N; i++) {
		int target = arr[i];
		ll t = 0;
		for (int j = 0; j < N; j++) {
			t += ((target - arr[j]) * (target - arr[j]));
		}
		ans = min(ans ,t);
	}
	cout << ans << endl;
	// string input;
	// vector<char> output;
	// cin >> input;
	// for (int i = 0; i < input.length(); i++) {
	// 	if (input[i] == '0') output.pb('0');
	// 	else if (input[i] == '1') output.pb('1');
	// 	else if (output.size()) output.pop_back(); 
	// }
	// for (int i = 0; i < output.size(); i++) cout << output[i];
	return 0;	
}