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, fastout};
use std::cmp::max;

#[fastout]
fn main() {
    input! {
        n: usize,
        ax: [i64; n]
    }

    let mut dp = vec![vec![0; n+1]; n+1];

    let mut axi: Vec<_> = ax.iter().enumerate().collect();
    axi.sort_by_key(|a| a.1);
    axi.reverse();

    let mut ans = 0;
    for i in 1..=n {
        for j in 0..=i {
            let (index, ai) = axi[i-1];
            if j > 0 {
                dp[i][j] = 
                    max(
                        dp[i][j],
                        dp[i-1][j-1] + ai * (index as i64 - (j-1) as i64).abs()
                    );
            }
            let k = i - j;
            if k > 0 {
                dp[i][j] = 
                    max(
                        dp[i][j],
                        dp[i-1][j] + ai * (index as i64 - (n-k) as i64).abs()
                    );
            }
            ans = max(ans, dp[i][j]);
        }
    }

    println!("{}", ans);
}