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;

fn main() {
    input! {
        n: usize,
        a: [isize; n],
    }
    let mut dp = vec![vec![0; n + 1]; n + 1];
    for (i, &arg) in arg_sort_descending(&a).iter().enumerate() {
        for l in 0..=i {
            let r = n - (i - l) - 1;
            dp[i + 1][l + 1] =
                dp[i + 1][l + 1].max(dp[i][l] + a[arg] * (arg as isize - l as isize));
            dp[i + 1][l] = dp[i + 1][l].max(dp[i][l] + a[arg] * (r as isize - arg as isize));
        }
    }

    println!("{}", dp.last().unwrap().iter().max().unwrap());
}

fn arg_sort_descending<T: Ord>(a: &Vec<T>) -> Vec<usize> {
    let mut i: Vec<_> = (0..a.len()).collect();
    i.sort_by(|&x, &y| a[y].cmp(&a[x]));
    i
}