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 fi first
#define endl "\n"
#define se second
#define ll long long
//priority_queue<int, vector<int>, greater<int> > pq;
const int N = 300030;
#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
const ll mod =    1e9+7;
int main()
{
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

int n;
cin>>n;
int arr[n+1];
rep(i,1,n+1)
cin>>arr[i];
int b[n+1];
rep(i,0,n+1)
b[i]=0;
int cnt=0;
int a[n+1];
rep(i,0,n+1)
a[i]=0;
rep(i,n+1,1)
{
int x=arr[i]%2;
int y=0;
if((i)*2<=n)

y=a[(i)*2];

if(y==x)
{
	
a[i]=y;
	continue;
}
if(y==1)
{
	cout<<"-1";
	return 0;
}
else
{
	a[i]=1;
	b[i]=1;
   cnt++;
}


}
cout<<cnt<<endl;
rep(i,1,n+1)
{
	if(b[i]>0)
	cout<<i<<endl;
}


}