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)]
use std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use proconio::{input, marker::*};
use superslice::*;
use rustc_hash::FxHashSet;
fn main() {
    input! {
        n: usize,
        ai: [i64; n],
    }
    let mut ai: Vec<_> = ai.into_iter().enumerate().map(|(i,a)| (a, i as i64)).collect();
    ai.sort_unstable();
    ai.reverse();

    let mut dp = vec![vec![0i64; n+1]; n+1];
    for i in 0..n {
        for low in 0..=i {
            let high = i - low;
            let (score, pos) = ai[i];

            let low_dist = (pos - low as i64).abs();
            let high_dist = (n as i64 - 1 - high as i64 - pos).abs();

            let low_score = dp[low][high] + score * low_dist;
            let high_score = dp[low][high] + score * high_dist;

            dp[low+1][high] = max(dp[low+1][high], low_score);
            dp[low][high+1] = max(dp[low][high+1], high_score);
        }
    }
    let mut ans = 0;
    for i in 0..=n {
        ans = max(ans, dp[i][n-i]);
    }
    println!("{}", ans);
}