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(unused_imports)]
use std::cmp::*;
use proconio::*;

#[fastout]
fn main() {
   input! {
      n: usize,
      A: [i64; n],
   }
   let mut prev = A.iter().zip(0..n).collect::<Vec<_>>();
   prev.sort();
   let mut dp: Vec<i64> = (0..n+1).map(|_| 0).collect();
   for i in 0..n {
       let (a, pos) = prev[i];
       for l in 0..n-i {
           let r = l + i;
           dp[l] = max(dp[l] + a*(r as i64-pos as i64).abs(), dp[l+1] + a * (l as i64-pos as i64).abs());
       }
   }
   println!("{}", dp[0]);
}