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)

fn main() {
    proconio::input! {
        n: usize,
        a: [usize; n],
    }
    let mut dp = vec![vec![0;n+1];n+1];
    let mut b: Vec<_> = (0..n).collect();
    b.sort_by_key(|k| a[*k]);
    for (m,k) in b.into_iter().rev().enumerate() {
        let v = a[k];
        let f = |a,b| v*if a>b {a-b} else {b-a};
        for i in 0..m+2 {
            let j = n+i-m-1;
            let x = if i>0 {dp[i-1][j]+f(i-1,k)} else {0};
            let y = if j<n {dp[i][j+1]+f(j,k)} else {0};
            dp[i][j]=std::cmp::max(x,y);
        }
    }
    println!("{}",(0..=n).map(|i| dp[i][n-i]).max().unwrap());
}