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: Time Limit Exceeded)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };
    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };
    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

pub trait LexicalPermutation {
    fn next_permutation(&mut self) -> bool;
    fn prev_permutation(&mut self) -> bool;
}
impl<T> LexicalPermutation for [T]
where
    T: Ord,
{
    fn next_permutation(&mut self) -> bool {
        if self.len() < 2 {
            return false;
        }
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] >= self[i] {
            i -= 1;
        }
        if i == 0 {
            return false;
        }
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i - 1] {
            j -= 1;
        }
        self.swap(j, i - 1);
        self[i..].reverse();
        true
    }
    fn prev_permutation(&mut self) -> bool {
        if self.len() < 2 {
            return false;
        }
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] <= self[i] {
            i -= 1;
        }
        if i == 0 {
            return false;
        }
        self[i..].reverse();
        let mut j = self.len() - 1;
        while j >= i && self[j - 1] < self[i - 1] {
            j -= 1;
        }
        self.swap(i - 1, j);
        true
    }
}

fn main() {
    input! {
        n: usize,
        aa: [u64;n]
    }
    let mut v = vec![vec![0;n];n];
    for (i,a) in aa.iter().enumerate() {
        v[i][i] = *a;
    }
    for i in  0..n {
        for j in 0..i {
            v[i][j] = v[i][i] * ((i-j) as u64);
        }
        for j in i+1..n {
            v[i][j] = v[i][i] * ((j-i) as u64);
        }
    }
    for i in  0..n {
        v[i][i] = 0;
    }
    // for x in &v {
    //     println!("{:?}", x);
    // }
    let mut p = (0..n).collect::<Vec<_>>();
    let mut sum: u64 = p.iter().enumerate().map(|(i,j)| v[i][*j]).sum();
    while p.next_permutation() {
        sum = std::cmp::max(sum,
            p.iter().enumerate().map(|(i,j)| v[i][*j]).sum());
        // println!("{:?}  {}", p, sum);
    }
    println!("{}", sum);
}