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)

use proconio::input;
use std::cmp::max;

fn main() {
    input!{
        n: usize,
        a: [usize; n],
    }

    let mut v: Vec<(usize, usize)> = Vec::new();
    for (i, j) in a.iter().enumerate() {
        v.push((j.clone(), i));
    }
    v.sort_by(|x, y| (&y.0).cmp(&x.0));
    
    let mut d = vec![vec![0; n + 1]; n + 1];
    let f = |i: i64, j: i64| -> usize {
        ((i - j).abs()) as usize
    };
    for (i, &(act, pos)) in v.iter().enumerate() {
        for x in 0..i + 1 {
            let y = i - x;
            let m_x = f(pos as i64, x as i64);
            let m_y = f((n - y - 1) as i64, pos as i64);
            d[x + 1][y] = max(d[x + 1][y], d[x][y] + m_x * act);
            d[x][y + 1] = max(d[x][y + 1], d[x][y] + m_y * act);
        }
    }
    let ans = d.iter()
        .flat_map(|v| v).max().unwrap();
    println!("{}", ans);
}