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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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

int main(void)
{
	long int i, j;
	long int n;
	long int flg;
	long int count;
	long int sum;
	long int a[200001];
	long int x[200001] = {0};
	
	scanf("%ld", &n);
	
	count = 0;
	for(i = 1; i <= n; i++) {
		scanf("%ld", &a[i]);
		if( (i > (long int)sqrt(n)) && (a[i] == 1) ) {
			x[i] = 1;
			count++;
		}
	}
	
	for(i = (long int)sqrt(n); i > 0; i--) {
		sum = 0;
		for(j = i; j <= n; j = j + i) {
			sum += a[j];
		}
		if( ((sum % 2) == a[i]) && (a[i] == 1) ) {
			x[i] = 1;
			count++;
		} else if( ((sum % 2) != a[i]) && (a[i] == 0) ) {
			x[i] = 1;
			count++;
		} else if( ((sum % 2) != a[i]) && (a[i] == 1) ) {
			printf("-1\n");
		}
	}
	
	printf("%ld\n", count);
	
	for(i = 1; i <= n; i++) {
		if(x[i] == 1) {
			printf("%ld ", i);
		}
	}
	
	return 0;
}