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)

#![allow(dead_code, unused_imports, deprecated)]
use std::cmp::{max, min, Reverse};

use proconio::input;
use proconio::marker::*;
use proconio::fastout;

const INF: usize = 1_000_000_007;

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

  let mut a = a
    .iter()
    .enumerate()
    .collect::<Vec<_>>();
  a.sort_by_key(|&(i, &x)| Reverse(x));
  let mut dp = vec![vec![0i64; n + 1]; n + 1];
  dp[1][1] = a[0].0 as i64 * a[0].1;
  dp[1][0] = (n as i64 - 1 - a[0].0 as i64) * a[0].1;
  for k in 2..=n {
    for l in 0..=k {
      let r = k - l;
      
      let mut act = vec![];
      
      if l > 0 {
        act.push(dp[k - 1][l - 1] + ((l - 1) as i64 - a[k - 1].0 as i64).abs() * a[k - 1].1);
      }
      if l < k {
        act.push(dp[k - 1][l] + ((n - r) as i64 - a[k - 1].0 as i64).abs() * a[k - 1].1);
      }
      dp[k][l] = *act.iter().max().unwrap();
    }
  }
  
  println!("{}", dp[n].iter().max().unwrap());
}

fn main() {
  std::thread::Builder::new()
    .name("big stack size".into())
    .stack_size(32 * 1024 * 1024)
    .spawn(|| {
      solve();
    })
    .unwrap()
    .join()
    .unwrap();
}