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;
use std::cmp;

const INF:i64 = 1e15 as i64;
fn main() {
    input! {
        n:usize,
        a:[i64;n],
    };

    let mut dp = vec![vec![-INF;n+1];n+1];
    dp[0][0] = 0;
    let mut pa:Vec<(usize,i64)> = a.into_iter().enumerate().map(|(i,x)|(i,x)).collect();
    pa.sort_by_key(|x|{
        -x.1
    });
    let pa = pa;
    for i in 0..n {
        let pos:i64 = pa[i].0 as i64;
        let a = pa[i].1;
        for j in 0..(i+1) {
            let l:i64 = j as i64;
            let r:i64 = (n-1-(i-j)) as i64;
            dp[i+1][j] = cmp::max(dp[i+1][j],(r-pos).abs()*a+dp[i][j]);
            dp[i+1][j+1] = cmp::max(dp[i+1][j+1],(pos-l).abs()*a+dp[i][j]);
        }
    }
    // println!("{:?}",dp);
    let res = dp[n].iter().max().unwrap();
    println!("{}",res);
}