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, min};

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

    let mut v = Vec::new();

    for (i, a) in aa.iter().enumerate() {
        v.push((a, i));
    }

    v.sort();
    v.reverse();

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

    for i in 0..n {
        let (a, p) = v[i];
        let p = p as i64;
        for x in 0..=i {
            let y = i - x;
            let i = i as i64;
            dp[x + 1][y] = max(dp[x + 1][y], dp[x][y] + (p - x as i64).abs() * a);
            dp[x][y + 1] = max(
                dp[x][y + 1],
                dp[x][y] + (p - (n as i64 - 1 - y as i64)).abs() * a,
            );
        }
    }

    let mut ans = 0;
    for v in dp {
        for v in v {
            if ans < v {
                ans = v;
            }
        }
    }

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