Problem:
Score : 700 points

Problem Statement

We have N gemstones labeled 1 through N.

You can perform the following operation any number of times (possibly zero).

Select a positive integer x, and smash all the gems labeled with multiples of x.

Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan).
However, a_i may be negative, in which case you will be charged money.

By optimally performing the operation, how much yen can you earn?

Constraints

All input values are integers.

1 \leq N \leq 100

|a_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the maximum amount of money that can be earned.

Sample Input 1

6
1 2 -6 4 5 3

Sample Output 1

12

It is optimal to smash Gem 3 and 6.

Sample Input 2

6
100 -100 -100 -100 100 -100

Sample Output 2

200

Sample Input 3

5
-1 -2 -3 -4 -5

Sample Output 3

0

It is optimal to smash all the gems.

Sample Input 4

2
-1000 100000

Sample Output 4

99000

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

#[allow(unused_macros)]
macro_rules! invec {
    ( $t:ty ) => {{
        let mut s = String::new();
        match std::io::stdin().read_line(&mut s) {
            Ok(0) => Vec::<$t>::new(),
            Ok(n) => s.trim().split_whitespace().map(|s| s.parse::<$t>().unwrap()).collect::<Vec<$t>>(),
            Err(_) => Vec::<$t>::new(),
        }
    }}
}

#[allow(unused_macros)]
macro_rules! input {
    ( $($t:ty),* ) => {{
        let mut s = String::new();
        std::io::stdin().read_line(&mut s);
        let mut splits = s.trim().split_whitespace();
        ($(
            {
                 splits.next().unwrap().parse::<$t>().unwrap()
            },
        )*)
    }}
}

#[allow(unused_must_use)]
#[allow(unused_variables)]
fn solve() {
    let (n,) = input!(usize);
    let mut arr = invec!(i64);

    // let mut total = arr.iter().sum();

    for i in (0..n).rev() {
        if arr[i] < 0 {
            let mut diff = arr[i];
            let mut mul = 2;
            while (i + 1) * mul <= n {
                diff += arr[(i + 1) * mul - 1];
                mul += 1
            }
            mul = 1;
            if diff < 0 {
                while (i + 1) * mul <= n {
                    arr[(i + 1) * mul - 1] = 0;
                    mul += 1
                }
            }
        }
    }

    println!("{}", arr.iter().sum::<i64>());
}

fn main() {
    solve();
}