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: Wrong Answer)

use std::cmp::max;
use std::io::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let mut a = Vec::new();
    for i in 0..n {
        let b: i64 = itr.next().unwrap().parse().unwrap();
        a.push((b, i as i64));
    }
    a.sort();
    a.reverse();

    let mut dp = vec![vec![0; 2020]; 2020];
    for i in 0..n {
        for j in 0..i + 1 {
            let k = n - (i - j);
            dp[j + 1][k] = max(dp[j + 1][k], dp[j][k] + a[i].0 * (j as i64 - a[i].1).abs());
            dp[j][k - 1] = max(
                dp[j][k - 1],
                dp[j][k] + a[i].0 * (k as i64 - 1 - a[i].1).abs(),
            );
        }
    }

    let mut ans = 0;
    for i in 0..n + 1 {
        ans = max(ans, dp[i][n - i]);
    }
    println!("{}", ans);
}