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: Accepted)

#include <bits/stdc++.h>
using namespace std;
vector<long long>done(200005);
vector<long long>memo(200005);
vector<long long>data;
long long n;//ここら辺少しうーんって感じ
long long dp(long long p){
if (done.at(p)){
	return memo.at(p);
}
else{
	long long res;
  int f=p;
  if (p*2>n){
  res=data.at(p-1);
  }
  else{
    int e=0;
    p=2*f;
    while (p<=n){
      e+=dp(p);
      p+=f;
    }
   
  if (e%2==data.at(f-1)){
  res=0;  
  }
    else{
      
  res=1;
    }
  }
    memo.at(f)=res;
    done.at(f)=1;
  return res;
  }
}
  
  
  



int main(){
  cin>>n;
  data.resize(n);
  for (long long i=0;i<n;i++){
  cin >>data.at(i);
  }
  dp(1);
  
 
  long long r;
   for (long long i=1;i<n+1;i++){
     r+=memo.at(i);
   }
  cout <<r<<endl;
 for (long long i=1;i<n+1;i++){
   if (memo.at(i)){
   cout <<i<<" "; 
   }
    //前からp桁目の数の状況。
  }  
}