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 std::cmp::max;
use proconio::input;

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

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

    let mut solve = Solve {
        n,
        a,
    };
    solve.solve();
}

struct Solve {
    n: usize,
    a: Vec<(u64, usize)>,
}

impl Solve {
    fn solve(&mut self) {
        let n = self.n;
        let a = &self.a;
        // dbg!(a);

        // 大きい方からi個見て、左端からj個、右端からi - j個埋める
        let mut dp = vec![vec![0u64; n + 1]; n + 1];

        for i in 1..=n {
            for j in 0..=i {
                let (a_i, origin_i) = a[i - 1];
                if j > 0 {
                    dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + a_i * (origin_i as i64 - (j - 1) as i64).abs() as u64);
                }
                let k = i - j;
                if k > 0 {
                    dp[i][j] = max(dp[i][j], dp[i - 1][j] + a_i * (origin_i as i64 - (n - k) as i64).abs() as u64);
                }
            }
        }

        let mut ans = 0u64;
        for j in 0..=n {
            ans = max(ans, dp[n][j]);
        }
        println!("{}", ans);
    }
}