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;
 
int main() {
  int N; cin >> N;
  vector<int> a(N);
  for(int i = 0; i < N; ++i){
    cin >> a.at(i);
  }
  
  int cnt1 = 0;
  
  vector<int> b(N);
  for(int i = N-1; i >= 0; --i){
    if(N / (i+1) == 1){
      b.at(i) = a.at(i);
      if(a.at(i) == 1) cnt1++;
    }else{
      int cnt = 0;
      for(int j = 0; j < N/(i+1); j++){
        cnt += b.at((i+1) * (j+1) - 1);
      }
      if(cnt / 2 == a.at(i)){
        b.at(i) = 0;
      }else{
        b.at(i) = 1;
        cnt1++;
      }
    }
  }
  
  cout << cnt1 << endl;
  for(int i = 0; i < N; ++i){
    if(b.at(i) == 1) cout << i+1 << endl;
  }
}