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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
using namespace std;

int main()
{
    // Input
    int n;
    cin >> n;

    vector<int> vecA(n);
    for(int i=0; i<n; i++)  cin >> vecA.at(i);

    // Calc
    vector<int> vecB(n);
    for(int i=n; i>=1; i--){
        int mod = 0;
        for(int j=2; i*j<=n; j++)
            mod += vecB[i*j-1];
        mod = mod%2;

        vecB[i-1] = (vecA[i-1] != mod) ? 1 : 0;
    }

    vector<int> vecAns;
    for(int i=0; i<vecB.size(); i++){
        if(vecB[i] == 0) continue;
        vecAns.push_back(i+1);
    }

    // Output
    cout << vecAns.size() << endl;
    for(int i=0; i<vecAns.size(); i++)
        printf("%d%c", vecAns[i], i+1==vecAns.size() ? '\n' : ' ');

    return 0;
}