Problem:
Score : 500 points

Problem Statement

There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.

You can rearrange these children just one time in any order you like.

When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.

Find the maximum total happiness points the children can earn.

Constraints

2 \leq N \leq 2000

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_2 ... A_N

Output

Print the maximum total happiness points the children can earn.

Sample Input 1

4
1 3 4 2

Sample Output 1

20

If we move the 1-st child from the left to the 3-rd position from the left, the 2-nd child to the 4-th position, the 3-rd child to the 1-st position, and the 4-th child to the 2-nd position, the children earns 1 \times |1-3|+3 \times |2-4|+4 \times |3-1|+2 \times |4-2|=20 happiness points in total.

Sample Input 2

6
5 5 6 1 1 1

Sample Output 2

58

Sample Input 3

6
8 6 9 1 2 1

Sample Output 3

85

Solution in Rust (1.42.0) (judge verdict: Accepted)

macro_rules! input {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

use std::cmp::max;

fn main() {
    let stdin = std::io::stdin();
    let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
    let mut next = move || -> String{
        bytes
            .by_ref()
            .map(|r|r.unwrap() as char)
            .skip_while(|c|c.is_whitespace())
            .take_while(|c|!c.is_whitespace())
            .collect()
    };
    input! {next,
        n: usize,
        a: [usize; n],
    }
    let mut a: Vec<_> = a.into_iter().enumerate().map(|(i,v)| (v,i)).collect();
    a.sort();
    a.reverse();
    // dbg!(&a);
    // dp[i][j] is the best result after using i numbers and filling j on the left ((i-j) on the right)
    let mut dp = vec![vec![0; n+1]; n+1];
    for (i, (activity, old_i)) in a.into_iter().enumerate() {
        dp[i+1][0] = dp[i][0] + (((n-1-(i-0)) as i64) - old_i as i64).abs() as usize * activity;
        for left in 1..=i+1 {
            // dbg!(i, left, (n-1-(i+1-left)));
            dp[i+1][left] = max(
                if i >= left {
                    dp[i][left] + (((n-(i+1-left)) as i64) - old_i as i64).abs() as usize * activity
                } else {
                    0
                },
                dp[i][left-1] + (((left-1) as i64) - old_i as i64).abs() as usize * activity
            );
            // dbg!(dp[i][left] + (((n-1-(i+1-left)) as i64) - old_i as i64).abs() as usize * activity);
            // dbg!(dp[i][left-1] + (((left-1) as i64) - old_i as i64).abs() as usize * activity);
            // dbg!(dp[i+1][left]);
        }
        // println!("{:?}", &dp[i+1]);
    }
    println!("{}", dp[n].iter().max().unwrap());
}