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)

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<set>
#include<queue>
#include<string>
#include<sstream>
#include<iostream>
#include<sstream>
#include<stack>
#include<map>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<cstdio>
using namespace std;
int a[1000];
int main()
{
	int sum = 0;
	int n;
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &a[i]);
		sum += a[i];
	}
	int t1 = sum / n;
	int t2 = t1 + 1;
	int sum1 = 0; int sum2 = 0;
	for (int i = 0; i < n; i++)
	{
		sum1 += (a[i] - t1)*(a[i] - t1);
	}
	for (int i = 0; i < n; i++)
	{
		sum2 += (a[i] - t2)*(a[i] - t2);
	}
	int mins = min(sum1, sum2);
	printf("%d", mins);
}