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};

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;

#[fastout]
fn main() {
    input! {
        n: usize,
        mut a: [isize; n],
    }

    let b = a.iter();
    let mut c: Vec<(&isize, usize)> = b.enumerate().map(|(a, b)| (b, a)).collect();
    c.sort();
    let d: Vec<usize> = c.iter().rev().map(|(_, index)| *index).collect();

    let mut hash = vec![0; n];
    for i in 0..n {
        hash[i] = d[i];
    }

    let mut dp = vec![vec![0 as isize; n + 1]; n + 1];
    for d in 1..=n {
        let i = hash[d - 1];
        for l in 0..=d {
            let r = d - l;
            if l > 0 {
                dp[l][r] = dp[l - 1][r] + a[i] * ((l as isize - 1) - i as isize).abs() as isize;
            }
            if r > 0 {
                dp[l][r] = max(
                    dp[l][r],
                    dp[l][r - 1] + a[i] * ((n - r) as isize - i as isize).abs(),
                );
            }
        }
    }

    let mut ans = 0;
    for l in 0..=n {
        let r = n - l;
        ans = max(ans, dp[l][r])
    }
    println!("{}", ans);
}