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 std::io::{self, BufRead};
use std::cmp::max;

fn abs(x: usize, y: usize) -> usize {
  if x < y { y - x } else { x - y }
}

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let line = lines.next().unwrap().unwrap();
  let n: usize = line.parse().unwrap();
  let line = lines.next().unwrap().unwrap();
  let mut xs: Vec<(usize, usize)> = line.split(' ').map(|x| x.parse().unwrap()).enumerate().collect();
  xs.sort_by(|x, y| y.1.cmp(&x.1));
  let mut memo: Vec<Vec<usize>> = vec![vec![0; n+1]; n+1];
  for k in 1..=n {
    let (p, a) = xs[k-1];
    for i in 0..=k {
      let j = k - i;
      if i == 0 {
        memo[i][j] = memo[i][j-1] + a * abs(p, n-j);
      } else if j == 0 {
        memo[i][j] = memo[i-1][j] + a * abs(p, i-1);
      } else {
        memo[i][j] = max(memo[i][j-1] + a * abs(p, n-j), memo[i-1][j] + a * abs(p, i-1));
      }
    }
  }
  let ans = (0..=n).map(|k| memo[k][n-k]).max().unwrap();
  println!("{}", ans);
}