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: Wrong Answer)

use std::io::*;

fn read_int(s: &mut StdinLock) -> i64 {
    s.by_ref().bytes().map(|c| c.unwrap() as char)
        .take_while(|c| !c.is_whitespace())
        .skip_while(|c| c.is_whitespace())
        .fold(0, |a, x| (x as u8 - b'0') as i64 + a * 10)
}
fn read_line_split_int(s: &mut StdinLock, n: i64) -> Vec<i64> {
    (0..n).map(|_| read_int(s)).collect()
}
fn main(){
    let s = stdin();
    let mut s = s.lock();
    let n = read_int(&mut s);
    let a = read_line_split_int(&mut s, n);
    let mut ans: Vec<(i64, i64)> = Vec::new();
    let mut answer: i64 = 0;
    let mut left = 0;
    let mut right = n-1;
    for i in 0..(n as usize) {
        ans.push((a[i], i as i64));
    }
    ans.sort_by_key(|x| x.0);
    'outer: for i in (0..(n as usize)).rev() {
        if (ans[i].1 - left).abs() > (right - ans[i].1).abs() {
            answer += ans[i].0 * (ans[i].1 - left).abs();
            left += 1;
        } else if (ans[i].1 - left).abs() < (right - ans[i].1).abs() {
            answer += ans[i].0 * (right - ans[i].1).abs();
            right -= 1;
        } else {
            for j in (0..i).rev() {
                if (ans[j].1 - left).abs() > (right - ans[j].1).abs() {
                    answer += ans[i].0 * (right - ans[i].1).abs();
                    right -= 1;
                    continue 'outer;
                } else if (ans[j].1 - left).abs() < (right - ans[j].1).abs() {
                    answer += ans[i].0 * (ans[i].1 - left).abs();
                    left += 1;
                    continue 'outer;
                } 
            }
            answer += ans[i].0 * (ans[i].1 - left).abs();
            left += 1;
        }
        
    }
    println!("{}", answer);
}