Problem:
Score : 400 points

Problem Statement

You are given N integers; the i-th of them is A_i.
Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.

Constraints

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the maximum possible sum of the absolute differences between the adjacent elements after arranging the given integers in a row in any order you like.

Sample Input 1

5
6
8
1
2
3

Sample Output 1

21

When the integers are arranged as 3,8,1,6,2, the sum of the absolute differences between the adjacent elements is |3 - 8| + |8 - 1| + |1 - 6| + |6 - 2| = 21. This is the maximum possible sum.

Sample Input 2

6
3
1
4
1
5
9

Sample Output 2

25

Sample Input 3

3
5
5
1

Sample Output 3

8

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}
fn main(){
    let n:usize  =read();
    let mut an = vec![];
    for _ in 0..n{
        an.push(read::<i64>());
    }
    let mut ans:Vec<i64> = vec![0;n];
    an.sort();
    let mid = if n % 2 == 0  {(n/2) -1} else { n /2  };
    // println!("{}",mid );
    let mut sum = 0;
    ans[mid] = an[0];
    for j in 1..(n/2)+1{
        if j % 2 != 0 {
            ans[mid + j] = an[n-j];
            sum += (ans[mid + j -1] - an[n-j]).abs();
            if !(n % 2 == 0 && j == n/2) { 
                ans[mid - j] = an[n-j-1];
                sum += (ans[mid -j + 1] - an[n-j-1]).abs();
            }
        } else {
            ans[mid+j] = an[j-1];
            sum += (ans[mid+j-1] - an[j-1]).abs();
            if !(n % 2 == 0 && j == n/2) {
                ans[mid-j] = an[j];
                sum += (ans[mid - j + 1] - an[j]).abs();
            }
        }
    }
    // println!("{:?}",ans );
    println!("{}",sum);
    // println!( );
}