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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>

int compare(const void *a, const void *b)
{
  return *(int *)b - *(int *)a;
}

int main (int argc, char **argv)
{
  int n = 0;
  int m = 0;
  int *a = NULL;
  int *b = NULL;
  int i = 0;
  int j = 0;
  int k = 0;

  scanf("%d", &n);

  a = (int *)malloc(n * sizeof(int));
  b = (int *)malloc(n * sizeof(int));
  memset(b, 0, n * sizeof(int));

  for(i = 0; i < n; i++)
  {
    scanf("%d", &a[i]);
  }

  for(i = n - 1; i >= 0; i--)
  {
    int l = 0;

    for(j = 2; (((i + 1) * j) - 1) < n; j++)
    {
      if(b[((i + 1) * j) - 1])
      {
        l = (l + 1) & 1;
      }
    }

    if(l != a[i])
    {
      b[i] = 1;
      m++;
    }
  }

  printf("%d\n", m);
  for(i = 0; i < n; i++)
  {
    if(b[i])
    {
      printf("%d ", b[i]);
    }
  }
  printf("\n");

  free(a);
  free(b);

  return 0;
}