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 <string.h>

int main(void) {
    int n, i, j;
    scanf("%d", &n);
    int a[n + 1];
    int ball[n + 1];  /* i 番目に ball を入れたら ball[i] = 1 */
    int temp_ball_sum[ n / 2 + 1];
    for ( i = 1; i <= n ; i++ ) {
        scanf("%d", &a[i]);
        ball[i] = 0;
        temp_ball_sum[i] = 0;
    }
    int ball_sum = 0;
    for ( i = n; i > 1; i-- ) {
        /* 後半は a[i] が 1 なら入れないといけない */
        if ( i > n / 2 ) {
            if ( a[i] == 1 ) {
                ball[i] = 1;
                ball_sum++;
                for ( j = 2; j <= n / 2; j++ ) {
                    if ( i % j == 0 ) {
                        temp_ball_sum[j] += 1;
                    }
                }
            }
        }
        /* 前半はこれまでの値に応じて入れていく */
        else {
            if ( a[i] != temp_ball_sum[i] % 2 ) {
                ball[i] = 1;
                ball_sum++;
            }
        }
    }
    /* 一番目の箱に入れるかどうか */
    if ( a[1] != ball_sum % 2 ) {
        ball[1] = 1;
        ball_sum++;
    }
    printf("%d\n", ball_sum);
    if ( ball_sum != 0 ) {
        for ( i = 1; i <= n; i++ ) {
            if ( ball[i] == 1 ) {
                printf("%d ", i);
            }
        }
        printf("\n");
    }
    return 0;
}