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: Wrong Answer)

#include <iostream>
#include <vector>
#include <map>
#include <functional>
#include <cmath>
#include <algorithm>

using namespace std;

int main(){
    int N;
    cin >> N;

    vector<int> a(N);
    vector<int> b(N);

    for (int i = 0; i < N; i++)
    {
        cin >> a[i];
        b[i]=0;
    }

//    cerr << N << endl;
    for (int i = N-1; i >=0 ; i--)
    {
  //      cerr << i << endl;
        if(b[i]!=a[i]){
            b[i]=1-b[i];
            vector<int> ret;

            for(int j=1 ; j*j<=i+1 ; ++j)
            {
                if((i+1)%j == 0)
                {
                    ret.push_back(j);
                    if(j!=1 && j*j!=(i+1))
                    {
                        ret.push_back((i+1)/j);
                    }
                }
            }
            for(auto j:ret){
                if(j<i) b[j-1] = 1 - b[j-1];
            }
            // for (int j = 0; j < sqrt(i+1)-1; j++)
            // {
            //     if( ((i+1)%(j+1)) == 0 ){
            //         b[j] = 1 - b[j];
            //     }
            // }
        }
        // for (int j = 0; j <N ; j++)
        // {
        //     cerr << b[j] << " ";
        // }
        // cerr << endl;
    }

    int cnt =0;
    for (int i = 0; i <N ; i++)
    {
        if(b[i]==1){
            cnt++;
        }
    }
    cout << cnt << endl;
    for (int i = 0; i <N ; i++)
    {
        if(b[i]==1){
            cout << i+1 << " ";
        }
    }
    cout << endl;  
    return 0;
}