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)

extern crate proconio;
use proconio::input;

fn main() {
    input! {
        n: usize,
        a: [i64; n],
    }
    let mut a = a.iter().zip(1..=(n as i64)).collect::<Vec<_>>();
    a.sort_by(|x, y| y.0.cmp(&x.0));
    let mut dp = vec![vec![0; n + 1]; n + 1];
    for i in 0..n {
        for j in 0..n {
            if i + j < n {
                let v = a[i + j].0;
                let k = a[i + j].1;
                // k -> i + 1
                dp[i + 1][j] =
                    std::cmp::max(dp[i + 1][j], dp[i][j] + v * (k - (i + 1) as i64).abs());
                // k -> n - j
                dp[i][j + 1] =
                    std::cmp::max(dp[i][j + 1], dp[i][j] + v * (k - (n - j) as i64).abs());
            }
        }
    }
    let mut ans = 0;
    for i in 0..=n {
        ans = std::cmp::max(ans, dp[i][n - i]);
    }
    println!("{}", ans);
}