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)]
#![allow(unused_macros)]
#![allow(non_snake_case)]
#![allow(dead_code)]
#![allow(unused_variables)]
#![allow(unused_mut)]
#![allow(unused_assignments)]
use proconio::input;
use proconio::marker::Usize1;
use std::collections::*;
use std::cmp::*;
use std::f64::consts::*;

const MOD: u64 = 1000000007;
const INF: i64 = std::i64::MAX / 4;

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();
    AI.sort();
    AI.reverse();

    let mut dp = vec![vec![-INF; 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 * ((l as i64) - j as i64).abs()));
            dp[l][r + 1] = max(dp[l][r + 1], dp[l][r] + (a * (((N as i64) - 1 - (r as i64)) - j as i64).abs()));
        }
    }

    println!("{}", dp.iter().map(|d| d.iter().max().unwrap()).max().unwrap());
}