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 ll long long
#define fi first
#define se second
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define rep1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define rep11(i, n) for (int i = 1; i < (int)(n); ++i)
#define repo(i, o, n) for (int i = o; i < (int)(n); ++i)
#define repm(i, n) for (int i = (int)(n)-1; i >= 0; --i)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define pb(n) push_back(n)
#define mp make_pair
#define MOD 1002000007
#define INF LONG_LONG_MAX

int main(){
   int n;
   cin>>n;
   int a[n+1];
   int b[n+1];
   int su=0;
   rep1(i,n){
       cin>>a[i];
       b[i]=0;
   }
   for (int i=n; i>0; --i){
       int p=1;
       int co=0;
       while (i*p<=n){
           co+=a[i*p];
           p++;
       }
       if (co%2==1){
           b[i]=1;
           su++;
       }
       
   }
   if (su==0){
       cout<<0<<endl;
   }
   else{
        cout<<su<<endl;
        rep1(i,n){
            if (b[i]==1){
                cout<<i;
                cout<<" ";
            }
        }
        cout<<endl;
   }
}