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: Accepted)


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

#define YYDBG 0

#if YYDBG==1
#define PRINTF(...) printf(__VA_ARGS__);
#define FOR for
#define IF if
#define ELSE else
#else
#define PRINTF(...) 
#define FOR(...)
#define IF(...)
#define ELSE 
#endif

int inputs[200000];
int outputs[200000];
int N;

int main(void) {
    int result = 0;
    int i,j;
    int num;
    int counter;
    int ans;

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

    PRINTF("in  : ");
    FOR(i=0;i<N;i++) {
        PRINTF("%d ", inputs[i]);
    }
    PRINTF("\n");
    PRINTF("\n");


    ans = 0;
    for(i=N;i>0;i--) {
        
        counter = 0;
        for(j=1;j<=N/i;j++) {
            PRINTF("-%d- ", j);
            counter += outputs[(j*i)-1];
        }
        if(counter%2 != inputs[i-1]) {
            outputs[i-1] = 1;
            ans++;
        }

        PRINTF("%d:%d(%d)[%d] ", i, inputs[i-1], N/i, counter);
    }
    PRINTF("\n");

    PRINTF("out : ");
    FOR(i=0;i<N;i++) {
        PRINTF("%d ", outputs[i]);
    }
    PRINTF("\n");
    PRINTF("\n");
    

    printf("%d\n", ans);
    for(i=0;i<N;i++) {
        if (outputs[i] == 1) {
            printf("%d ", i+1);
        }
    }
    printf("\n");

    return 0;
}