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 Rust (1.15.1) (judge verdict: Accepted)

#![allow(unused)]

use std::cmp::{max, min};
use std::io::stdin;
use std::mem::swap;
use std::str::FromStr;

fn input<T: FromStr>() -> Result<T, T::Err> {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim().parse()
}

fn input_vec<T: FromStr>() -> Result<Vec<T>, T::Err> {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim()
        .split_whitespace()
        .map(|x| x.parse())
        .collect()
}

fn main() {
    let n: usize = input().unwrap();
    let a: Vec<i32> = input_vec().unwrap();

    let mut ans = std::i32::MAX;
    for base in -100..101 {
        ans = min(ans, a.iter().map(|x| (*x - base) * (*x - base)).sum());
    }

    println!("{}", ans);
}