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, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};

#[fastout]
fn main() {
    input! {
    n:usize,
    a:[i64;n],
     }
    let mut ai = vec![(0, 0); n];
    for (i, &x) in a.iter().enumerate() {
        ai[i] = (x, i);
    }
    ai.sort_by_key(|x| -x.0);
    let mut dp = vec![vec![-INF; n + 1]; n + 1];
    dp[0][0] = 0;
    for i in 0..n {
        let k = ai[i].0;
        let r = ai[i].1;
        for x in 0..=i {
            let y = i - x;
            dp[x + 1][y] = max(dp[x + 1][y], dp[x][y] + k * (x as i64 - r as i64).abs());
            dp[x][y + 1] = max(
                dp[x][y + 1],
                dp[x][y] + k * ((n - 1 - y) as i64 - r as i64).abs(),
            );
        }
    }
    // dbg!(&dp);
    let mut ans = 0;
    for i in 0..=n {
        ans = max(ans, dp[i][n - i]);
    }
    println!("{}", ans);
}
const INF: i64 = 1 << 60;