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;
#[allow(unused_imports)]
use proconio::marker::*;

fn main() {
    input! {
        n: usize,
        a: [i64; n],
    }
    let mut a = a
        .iter()
        .enumerate()
        .map(|(i, x)| (i as i64, *x))
        .collect::<Vec<_>>();
    a.sort_by_key(|x| -x.1);

    let mut dp = vec![vec![0; n + 1]; n + 1];
    for i in 0..n {
        let pi = a[i].0;
        for l in 0..=i {
            let r = i - l;
            use std::cmp::max;
            dp[i + 1][l + 1] = max(dp[i + 1][l + 1], dp[i][l] + (pi - l as i64) * a[i].1);
            dp[i + 1][l] = max(dp[i + 1][l], dp[i][l] + ((n - 1 - r) as i64 - pi) * a[i].1);
        }
    }

    println!("{}", dp[n].iter().max().unwrap());
}