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};

use std::cmp::max;

#[fastout]
fn main() {
    input! {
        n: usize,
        a: [u64; n]
    }
    let mut ans = 0;
    let mut dp = vec![vec![0; n + 1]; n + 1];
    let mut a = a.iter().enumerate().collect::<Vec<_>>();
    a.sort_by_key(|x| std::cmp::Reverse(x.1));
    let abs = |x, y| if x > y { x - y } else { y - x };
    for i in 0..n {
        for j in 0..=i {
            let k = i - j;
            dp[i + 1][j] = max(
                dp[i + 1][j],
                dp[i][j] + a[i].1 * abs(n - 1 - k, a[i].0) as u64,
            );
            dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + a[i].1 * abs(a[i].0, j) as u64);
        }
    }
    for x in dp[n].iter() {
        ans = max(ans, *x);
    }
    println!("{}", ans);
}