Problem:
Score : 400 points

Problem Statement

There are N empty boxes arranged in a row from left to right.
The integer i is written on the i-th box from the left (1 \leq i \leq N).

For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it.

We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied:

For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2.

Does there exist a good set of choices? If the answer is yes, find one good set of choices.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

a_i is 0 or 1.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

If a good set of choices does not exist, print -1.

If a good set of choices exists, print one such set of choices in the following format:

M
b_1 b_2 ... b_M

where M denotes the number of boxes that will contain a ball, and b_1,\ b_2,\ ...,\ b_M are the integers written on these boxes, in any order.

Sample Input 1

3
1 0 0

Sample Output 1

1
1

Consider putting a ball only in the box with 1 written on it.

There are three boxes with multiples of 1 written on them: the boxes with 1, 2, and 3. The total number of balls contained in these boxes is 1.

There is only one box with a multiple of 2 written on it: the box with 2. The total number of balls contained in these boxes is 0.

There is only one box with a multiple of 3 written on it: the box with 3. The total number of balls contained in these boxes is 0.

Thus, the condition is satisfied, so this set of choices is good.

Sample Input 2

5
0 0 0 0 0

Sample Output 2

0

Putting nothing in the boxes can be a good set of choices.

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

use std::io::*;
use std::str::*;

fn read<T: FromStr>(s: &mut StdinLock) -> Option<T> {
    let s = s
        .by_ref()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok()
}

fn divisors(n: usize) -> Vec<usize> {
    let mut divisors = Vec::new();
    // n := i * x とおくと、 i が i > root(n) の時、　i はすでに ある x に探索されているから
    // i <= root(n) まで探索すればよい
    for i in 1..(f64::sqrt(n as f64) + 1e-9) as usize {
        // i で n が割り切れた場合
        if n % i == 0 {
            // 約数リストに格納
            divisors.push(i);

            // n := i * x の x を格納。ただし x := i の時は除く
            if i != n / i {
                divisors.push(n / i);
            }
        }
    }
    divisors.sort();
    divisors
}

fn main() {
    let stdin = std::io::stdin();
    let n = read::<u32>(&mut stdin.lock()).unwrap();

    let mut a = [0u32; 200000];

    let mut count = 0;
    let mut is = Vec::new();
    for i in 0..n {
        unsafe {
            let an = read::<u32>(&mut stdin.lock()).unwrap();
            if an == 1 {
                let mut divs = divisors(i as usize);
                divs.push(i as usize);
                for j in divs {
                    *a.get_unchecked_mut(j) = *a.get_unchecked(j) ^ 1;
                }
                count += 1;
                is.push(i + 1);
            };
        }
    }
    println!("{}", count);
    for i in is {
        print!("{} ", i);
    }
    println!("");
}