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::Reverse;

fn main() {
   input! {
      n:usize,
      a:[usize;n]
   }
   let mut v = vec![];
   for (i, x) in a.iter().enumerate() {
      v.push((x, i));
   }
   v.sort_by_key(|&x| Reverse(x.0));
   let mut dp = vec![vec![0; n + 2]; n + 2];
   for i in 0..n {
      for j in 0..n {
         if i + j >= n {
            continue;
         }
         dp[i][j + 1] = std::cmp::max(
            dp[i][j + 1],
            dp[i][j]
               + (n as isize - j as isize - 1 - v[i + j].1 as isize).abs() as usize * v[i + j].0,
         );
         dp[i + 1][j] = std::cmp::max(
            dp[i + 1][j],
            dp[i][j] + (i as isize - v[i + j].1 as isize).abs() as usize * v[i + j].0,
         );
      }
   }
   let mut ans = 0;
   for i in 0..=n {
      ans = std::cmp::max(ans, dp[i][n - i]);
   }
   println!("{}", ans);
}