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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define N 200010
int n, a[N], b[N];

bool ok() {
    for (int i = 1; i <= n; ++i) {
        int tot = 0;
        for (int j = i; j <= n; j += i) {
            tot = tot + b[j];
            tot %= 2;
        }
        if (tot ^ a[i]) return 0;
    }
    return 1;
}

int main() {
    while (scanf("%d", &n) != EOF) {
        for (int i = 1; i <= n; ++i) scanf("%d", a + i);
        for (int i = n; i >= 1; --i) {
            int now = 0;
            for (int j = i + i; j <= n; j += i) {
                now += b[j];
                now %= 2;
            }
            b[i] = now ^ a[i];
        }

            vector <int> vec;
            for (int i = 1; i <= n; ++i) if (b[i]) vec.push_back(i);
            int sze = (int)vec.size();
            printf("%d\n", sze);
            for (int i = 0; i < sze; ++i) printf("%d%c", vec[i], " \n"[i == sze - 1]);

    }
    return 0;
}