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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <random>
#include <bitset>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
typedef long long ll;
using namespace std;
const ll MOD = 1000000007LL;

// 約数のvectorを返す
vector<int> divisor(int n) {
    vector<int> ret;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            ret.push_back(i);
            if (i != n / i) ret.push_back(n / i);
        }
    }
    return ret;
}

int main() {
    cin.sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin >> n;

    vector <int> a(n+1, 0);
    vector <vector<int>> div(n+1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        div[i] = divisor(n);
    }

    vector <int> ans;
    for (int i = n; i > 0; i--) {
        if (a[i]) {
            ans.push_back(i);
            for (int d : div[i]) {
                a[d] = (a[d] + 1) % 2;
            }
        }
    }

    cout << ans.size() << "\n";
    for (int ret : ans) cout << ret << " ";

    return 0;
}