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 <iostream>
#include <vector>
#include <cstdio>
#include <string>
#include <algorithm>
#include <map>
#include <numeric>
using namespace std;


int main()
{
    int n;
    cin >> n;
    vector<int> a;
    for (int i = 0; i< n; i++) {
        int tmp;
        cin >> tmp;
        a.push_back(tmp);
    }

    int index = n / 2;
    vector<int> res(n);
    copy(a.begin(), a.end(), res.begin());

    for (int i = index; i > 0; i--) {
        int sum = 0;
        for (int j = n / i ; j > 1; j--) {
            sum += res[i * j - 1];
        }
        res[i - 1] = (sum % 2 + a[i - 1] == 1) ? 1 : 0;
    }
    int m = accumulate(res.begin(), res.end(), 0);
    cout << m << endl;
    for (int k = 0; k < n; k++) {
        if (res[k] == 1) {
            cout << k + 1 << endl;
        }
    }
}