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 proconio::*;

fn run() {
    input! {
        n: usize,
        a: [i64; n],
    }
    let mut ans = 0;
    for i in 0..n {
        let mut sum = 0;
        let mut f = vec![];
        for j in 0..i {
            sum -= a[j] * j as i64;
            f.push(a[j]);
        }
        let mut t = vec![];
        for j in i..n {
            sum += a[j] * j as i64;
            t.push(a[j]);
        }
        f.sort();
        t.sort();
        for j in ((n - i)..n).rev() {
            sum += f.pop().unwrap() * j as i64;
        }
        for j in 0..(n - i) {
            sum -= t.pop().unwrap() * j as i64;
        }
        ans = std::cmp::max(ans, sum);
    }
    println!("{}", ans);
}

fn main() {
    run();
}