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::{fastout, input};

#[fastout]
fn main() {
    input! {
        n: usize,
        a: [u64; n],
    }
    let mut dp = vec![vec![0 as u64; n + 1]; n + 1];
    let mut v = vec![0; n];
    for i in 0..n {
        v[i] = i;
    }
    let v = v.as_mut_slice();
    v.sort_by(|&i, &j| {
        if a[i] != a[j] {
            return a[i].cmp(&a[j]).reverse();
        } else {
            return i.max(n - 1 - i).cmp(&j.max(n - 1 - j));
        }
    });

    for i in 0..n {
        let j = v[i];
        for l in 0..i + 1 {
            let r = i - l;
            let r = n - 1 - r;
            {
                let d = j.max(l) - j.min(l);
                let tmp = (d as u64) * a[j];
                dp[i + 1][l + 1] = dp[i + 1][l + 1].max(tmp + dp[i][l]);
            }
            {
                let d = j.max(r) - j.min(r);
                let tmp = (d as u64) * a[j];
                dp[i + 1][l] = dp[i + 1][l].max(tmp + dp[i][l]);
            }
        }
    }

    let ans = dp[n].iter().fold(0 as u64, |res, &x| res.max(x));
    println!("{}", ans);
}