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 JavaScript (node.js v5.12) (judge verdict: Accepted)

"use strict";
    
const main = arg => {
    arg = arg.trim().split("\n");
    const N = parseInt(arg[0]);
    const A = arg[1].split(" ").map(n=>parseInt(n));
    
    // 逆から考える
    A.reverse();
    
    let answer = [];
    let list = [];
    
    for(let i in A) {
        let alreadyIn = 0;
        
        for(let j=i-(N - i); j>=0; j -= (N - i)) {
            if(list[j] === 1) {
                alreadyIn++;
            }
        }

        if(alreadyIn % 2 === 0) {
            if(A[i] !== 0) {
                answer.push(N - i);
                list.push(1);
            } else {
                list.push(0);
            }
        } else {
            if(A[i] === 0) {
                answer.push(N - i);
                list.push(1);
            } else {
                list.push(0);
            }
        }
    }

    const cnt = answer.length;
    answer.sort((a,b)=>a-b);

    console.log(cnt);
    console.log(answer.join("\n"));
}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));