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::*;

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

    let mut ai: Vec<(i64, i64)> = a.iter().enumerate().map(|(i, &a)| (a, i as i64)).collect();
    //    let mut ai: Vec<(i64, i64)> = a.iter().enumerate().map(|(i, &a)| (a, i as i64)).collect();
    ai.sort();
    ai.reverse();
    // 降順に並んだ。
    // println!("{:?}", ai);
    //for (a, i) in &ai {
    //    println!("{},{}", a, i);
    //}
    let minimum = 0;
    let mut dp = vec![vec![minimum; n + 1]; n + 1];
    dp[0][0] = 0;

    for (i, &(a, j)) in ai.iter().enumerate() {
        for l in 0..i + 1 {
            let r = i - l;
            dp[l + 1][r] = max(dp[l + 1][r], dp[l][r] + a * (j - (l as i64)));
            dp[l][r + 1] = max(dp[l][r + 1], dp[l][r] + a * ((n - 1 - r) as i64 - j));
        }
    }
    //let mut ans = 0;    for l in 0..n {        let r = n - l;        ans = max(dp[l][r], ans);}
    //println!("{}", ans);
    println!(
        "{}",
        dp.iter().map(|d| d.iter().max().unwrap()).max().unwrap()
    );
}