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: Runtime Error)

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cstdio>
#include<cmath>
#include<numeric>
#include<queue>
#include<stack>
#include<cstring>
#include<limits>
#define rep(i,a) for(int i=(int)0;i<(int)a;++i)
#define pb push_back
#define eb emplace_back
#define mpr make_pair
#define fi first
#define se second
using ll=long long int;
static const ll mod = 1e9 + 7;
static const ll INF = 1LL << 50;
using namespace std;

int main(){
	ll n;
	cin>>n;
	vector<ll>a(n),b(n);
	rep(i,n){
		cin>>a[i];
		if((i+1)>n/2)b[i]=a[i];
	}
	for(int i=n/2;i>=0;--i){
		ll sum=0;
		for(int j=2;;++j){
			if(j>n)break;
			sum+=a[i*j];
		}
		ll jd=(2-sum%2)%2;
		 b[i]=1-jd;
	}
	ll cnt=0;
	rep(i,n){
		if(b[i]!=0)++cnt;
	}
	cout<<cnt<<endl;
	rep(i,n){
		if(b[i]!=0){
			cout<<i+1<<endl;
		}
	}
		return 0;
}