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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
 
int main() {
  int N;
  cin >> N;
  int A[N], B[N];
  for(int i = 0; i < N; i++) {
    B[i] = 0;
  }
  for(int i = 0; i < N; i++) {
    cin >> A[i];
  }
  bool ans = false;
  for(int i = N - 1; i >= 0; i--) {
    if(B[i] % 2 == 0) B[i] = A[i];
    if(B[i] % 2 == 1) {
      if(A[i] == 1) B[i] = 0;
      else B[i] = 1; 
    }
    if(A[i] == 1) {
      for(int j = 0; j * j <= i; j++) {
        if(i % j == 0) {
          B[j]++;
        }
      }
    }
  }
  int sum = 0;
  for(int i = 0; i < N; i++) {
    sum += B[i];
  }
  cout << sum << endl;
  for(int i = 0; i < N; i++) {
    if(B[i] == 1) cout << i + 1 << endl;
  }
}