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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define _GLIBCXX_DEBUG
#define rep(i, from, to) for (int i = from; i < (to); ++i)
#define mp(x,y) make_pair(x,y)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
using ll = long long;
using vin=vector<int>;
using vll=vector<ll>;
using P = pair<int, int>;
const int inf=1e9+7;
const ll INF=1e18;
template <typename T> void chmin(T &a, T b) { a = min(a, b); }
template <typename T> void chmax(T &a, T b) { a = max(a, b); }

int main() {
  int n;
  cin>>n;
  vin a(n);
  rep(i,0,n) cin>>a[i];
  int m=0;
  vin b(n,0);
  
  for(int i=n-1;i>=0;i--){
                int cnt=0;//iの倍数のはこに何個入っているか mod2
                int s=i;
                
    
                while(s+i+1<=n-1){
                          s+=(i+1);
                          if(b[s]==1){
                            cnt+=1;
                            cnt%=2;
                           // cout<<s<<a[i]<<cnt<<endl;
                          }
                }
   
    
               if(a[i]!=cnt){
                           m++;
                           b[i]=1;
               }
  }
  cout<<m<<endl;
  rep(i,0,m){
    if(b[i]==1)
   cout<<i+1<<endl; 
  }
}