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 <algorithm>
#include <climits>
#include <complex>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>

// using
using namespace std;

// typedef
typedef long long ll;

// define
#define FOR(i, a, b, c) for (int i = (int)(a); i < (int)(b); i += (int)(c))
#define REP(i, n) FOR(i, 0, n, 1)
#define RFOR(i, a, b, c) for (int i = (int)(a); i >= (int)(b); i -= (int)(c))
#define RREP(i, n) RFOR(i, n, 0, 1)
#define ALL(c) (c).begin(), (c).end()
#define SORT(c) sort(ALL(c))
#define REVERSE(c) reverse(ALL(c))
#define UNIQ(c) unique(ALL(c))
#define LB(c, x) lower_bound(c.begin(), c.end(), x)
#define UB(c, x) upper_bound(c.begin(), c.end(), x)
#define LI(c, x) distance(c.begin(), LB(c, x))
#define UI(c, x) distance(c.begin(), UB(c, x))

// functions
template <class T> T ceil(T a, T b) {
    return (a + b - 1) / b;
}
template <class T> T round(T a, T b) {
    return (a + b / 2) / b;
}
template <class T> T gcd(T a, T b) {
    return b ? gcd(b, a % b) : a;
}
template <class T> T lcm(T a, T b) {
    return a / gcd(a, b) * b;
}
template <class T> bool amax(T& a, const T& b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> bool amin(T& a, const T& b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}
ll modpow(ll a, ll b, ll m) {
    if (b == 0) {
        return 1;
    } else if (b % 2 == 0) {
        ll h = modpow(a, b / 2, m);
        return h * h % m;
    } else {
        return a * modpow(a, b - 1, m) % m;
    }
}
ll comb(ll a, ll b, ll m) {
    if (b > a - b) {
        return comb(a, a - b, m);
    }
    ll c = 1, d = 1;
    REP(i, b) {
        c *= (a - i);
        d *= (b - i);
        c %= m;
        d %= m;
    }
    return c * modpow(d, m - 2, m) % m;
}

// main
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    cin >> N;
    vector<int> a(N + 1);
    vector<bool> b(N + 1, false);
    FOR(i, 1, N + 1, 1) {
        cin >> a[i];
        if (i > N / 2) {
            b[i] = a[i] == 1;
        }
    }
    RFOR(i, N / 2, 1, 1) {
        int cnt = 0;
        FOR(j, 1, N / i + 1, 1) {
            if (b[i * j]) {
                cnt++;
            }
        }
        if (cnt % 2 != a[i]) {
            b[i] = true;
        }
    }
    int res = 0;
    FOR(i, 1, N + 1, 1) {
        if (b[i]) {
            res++;
        }
    }
    cout << res << endl;
    FOR(i, 1, N + 1, 1) {
        if (b[i]) {
            cout << i << " ";
        }
    }
    cout << endl;
    return 0;
}