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)

#define _USE_MATH_DEFINES
#include <iomanip>
#include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <tuple>
#include <climits>
#include <algorithm>
#include <cmath>
using namespace std;

typedef long long LL;
typedef vector<int> VI;
typedef vector<double> VD;
typedef vector<LL> VLL;
typedef vector<string> VS;
typedef vector<bool> VB;
typedef vector<vector<int>> VVI;
typedef vector<vector<bool>> VVB;
typedef vector<vector<double>> VVD;

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define all(a) (a).begin(),(a).end()

template<class T>bool chmax(T &a, const T &b){
	if(a<b){
		a=b;
		return true;
	}
	return false;
}

template<class T>bool chmin(T &a, const T &b){
	if(a>b){
		a=b;
		return true;
	}
	return false;
}

int main(){
	int n;cin>>n;
	VI a(n);
	rep(i,n){
		cin>>a.at(i);
	}
	int b = 0;
	for(int i=0;i<pow(2,n);i++){
		bool ok = true;
		for(int j=0;j<n;j++){
			int res = 0;
			for(int k=j;k<n;k++){
				if((k+1)%(j+1)==0){
					res += b & (int)(pow(2,j));
				}
			}
			res %= 2;
			if(res != a.at(j)){
				ok=false;
				break;
			}
		}
		if(ok){
			break;
		}else{
			b++;
		}
		if(i == pow(2,n)-1){
			cout<<-1<<endl;
			return 0;
		}
	}
	int num =0;
	VI ans;
	int box=1;
	while(b>0){
		num += b%2;
		if(b%2){
			ans.push_back(box);
		}
		box++;
		b/=2;
	}
	if(!num)ans.push_back(0);
	reverse(all(ans));
	cout<<num<<endl;
	for(int i:ans)cout<<i<<" ";
	cout<<endl;
	return 0;
}