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;

macro_rules! chmax {
    ($l: expr, $r: expr) => {
        $l = std::cmp::max($l, $r);
    };
}

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

    let mut pv: Vec<(i64, i64)> = av.iter().enumerate().map(|(i, &a)| (a, i as i64)).collect();
    pv.sort_by_key(|(a, i)| (-a, -i));
    // dp[i][l] ... 降順で i 番目まで見て l 個を左側に配置したときの最大値
    let mut dp = vec![vec![0_i64; n + 1]; n + 1];
    for i in 0..n {
        let (a, i_a) = pv[i];
        for l in 0..i + 1 {
            let r = n - (i - l) - 1; // (i - l) 個が既に右側に配置されている
            chmax!(dp[i + 1][l + 1], dp[i][l] + a * (i_a - l as i64).abs());
            chmax!(dp[i + 1][l], dp[i][l] + a * (i_a - r as i64).abs());
        }
    }

    let ans = dp[n].iter().max().unwrap();
    println!("{}", ans);
}