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)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use proconio::{fastout, input};
use std::cmp::max;

#[fastout]
fn main() {
    input! {
        n: usize,
        a: [u64; n],
    }

    let a = {
        let mut a = a
            .iter()
            .enumerate()
            .map(|(i, &x)| (x, i))
            .collect::<Vec<(u64, usize)>>();
        a.sort();
        a.reverse();
        a
    };

    let mut dp = vec![vec![0u64; n + 1]; n + 1];
    let dist = |a: usize, b: usize| (a as i64 - b as i64).abs() as u64;
    for u in 0..n {
        let (x, i) = a[u];
        for lef in 0..u + 1 {
            let rig = n - 1 - (u - lef);
            dp[u + 1][lef + 1] = max(dp[u + 1][lef + 1], dp[u][lef] + x * dist(i, lef));
            dp[u + 1][lef] = max(dp[u + 1][lef], dp[u][lef] + x * dist(i, rig));
        }
    }

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

*/

fn main() {
    let exe = "./binXXXXXX";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";