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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

int compare(const void* a, const void* b){
    return *(int*)a - *(int*)b;
}

int MIN(int a, int b){return a>b?b:a;}

int main(void){
    int n;
    scanf("%d", &n);
    int a[n];
    for(int i=0; i<n; i++){scanf("%d", &a[i]);}

    qsort(a, n, sizeof(int), compare);

    int min = a[0];
    int max = a[n-1];

    int sum = 0;
    int ans = 10000;
    for(int i=min; i<=max; i++){
        sum = 0;
        for(int j=0; j<n; j++){
            sum += (i-a[j])*(i-a[j]);
        }
        ans = MIN(ans,sum);
    }

    printf("%d\n", ans);
    return 0;
}