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>

void func(int* p, int* b, int x){
    int y = x;
    *(p+x) += 1;
    *(b+x) = (*(p+x)) % 2;
    if(y != 1){
        *(p+1) += 1;
        *(b+1) = (*(p+1)) % 2;
    }
    if(x % 2 == 0 && y != 2){
        *(p+2) += 1;
        *(b+2) = (*(p+2)) % 2;
        while(x % 2 == 0){
            x = x/2;
        }
    }
    for(int i=3; i*i<=x; i+=2){
        if(x % i == 0){
            *(p+i) += 1;
            *(b+i) = (*(p+i)) % 2;
            while(x % i == 0){
                x = x/i;
            }
        }
    }
    if(x != 1 && x != y){
        *(p+x) += 1;
        *(b+x) = (*(p+x)) % 2;
    }
}

int main(){
    int n;
    scanf("%d", &n);
    int a[n+1];
    for(int i=1; i<=n; i++){
        scanf("%d", &a[i]);
    }
    int b[n+1];
    int sum[n+1];
    int c=0;
    for(int i=0; i<=n; i++){
        b[i] = 0;
        sum[i] = 0;
    }
    for(int i=n; i>=1; i--){
        if(b[i] != a[i]){
            c++;
            func(sum, b, i);
        }
    }
    int f=1;
    for(int i=1; i<=n; i++){
        if(b[i] != a[i]){
            f = 0;
        }
    }
    if(f){
        printf("%d\n", c);
        for(int i=1; i<=n; i++){
            if(b[i]){
                printf("%d ", i);
            }
        }
    }else{
        printf("-1");
    }

    return 0;
}