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 <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define watch(x) cout<<"->"<<#x<<" : "<<x<<endl;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c)) 
#define all(v) (v).begin(),(v).end()
typedef long long ll;
typedef unsigned long long llu;
const double PI = 2*acos(0.0);
#define int long long
signed main()
{
	//freopen("","r",stdin);
	//freopen("","w",stdout);
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n;
	cin>>n;
	int a[n];
	for(int i=0;i<n;i++)cin>>a[i];
	vector<int> res;
	bool v = 1;
	for(int i=1;i<=n;i++)
	{
		int sum = a[i-1];
		res.push_back(a[i-1]);
		for(int j=i;j<=n;j+=i)
			if(i!=j)
				sum+=a[j-1];
		if(sum%2!=a[i-1])
		{
			v=0;
			break;
		}
	}	
	if(v)
	{
		cout<<res.size()<<endl;
		for(int val : res)
			cout<<val<<" ";
	}
	else cout<<"-1"<<endl;
	return 0;
}