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;

fn main() {
  input! {
    n: usize,
    a: [i64; n],
  }
  let mut b: Vec<(i64, usize)> = vec![(0, 0); n];
  for (i, &val) in a.iter().enumerate() {
    b[i] = (val, i);
  }
  b.sort(); b.reverse();
  let mut dp = vec![vec![0i64; n+1]; n+1];

  for i in 0..n {
    for j in 0..i+1 {
      dp[i+1][j] = std::cmp::max(dp[i+1][j], dp[i][j] + b[i].0 * ((b[i].1 as i64 - (n - (i-j) - 1) as i64) as i64).abs());
      dp[i+1][j+1] = std::cmp::max(dp[i+1][j+1], dp[i][j] + b[i].0 * ((b[i].1 as i64 - j as i64)).abs());
    }
  }
  println!("{}", dp[n].iter().fold(0i64, |x, &y| std::cmp::max(x, y)));
}