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::{input,fastout};
use std::cmp::{max, min};
#[fastout]
fn main(){
    input!{
        n: usize,
        a: [i64;n]
    }
    let mut b: Vec<_> = a.into_iter().enumerate().collect();
    b.sort_by_key(|p| p.1);
    let diff = |a:usize, b:usize| -> i64 {
        (max(a, b) - min(a, b)) as i64
    };
    let mut dp = vec![0];
    while let Some((k, a)) = b.pop() {
        let len = dp.len();
        let mut next = vec![0; len + 1];
        for (i, v) in dp.into_iter().enumerate() {
            next[i] = max(next[i], v + a * diff(k, n - (len - i)));
            next[i+1] = max(next[i + 1], v + a * diff(k, i));
        }
        dp = next;
    }
    let ans = dp.into_iter().max().unwrap();
    println!("{}", ans);
}