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: Wrong Answer)

use proconio::input;

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

  let mut v: Vec<(i64, i64, usize, usize)> = Vec::with_capacity(n*n);
  let mut occupied = vec![false; n];
  let mut used = vec![false; n];
  for i in 0..n {
    for j in 0..n {
      v.push((a[i] * (i as i64 - j as i64).abs(), -a[i],  i, j));
    }
  }
  v.sort();
  v.reverse();
  let mut res: i64 = 0;
  for (cost, _, i, j) in v {
    if !used[i] && !occupied[j] {
      res += cost;
      used[i] = true;
      occupied[j] = true;
    }
  }
  println!("{}", res);
}