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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.Arrays;
import java.util.Scanner;

public class Main {
  Scanner sc = new Scanner(System.in);

  public static void main(String[] args) {
    new Main().run();
  }

  void run() {
    int n = ni();
    int[] a = new int[n];
    for (int i = 0; i < n; ++i) {
      a[i] = ni();
    }

    int min = 1 << 28;
    for (int i = -100; i <= 100; ++i) {
      int sum = 0;
      for (int j = 0; j < n; ++j) {
        sum += (a[i] - i) * (a[i] - i);
      }
      min = Math.min(min, sum);
    }
    System.out.println(min);
  }

  int ni() {
    return Integer.parseInt(sc.next());
  }

  void debug(Object... os) {
    System.err.println(Arrays.deepToString(os));
  }
}