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 <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <stack>
#include <numeric>
#include <algorithm>
#include <stdio.h>
#include <bitset>
#include <complex>
#include <iomanip>
#include <math.h>
using namespace std;
#define MM 1000000000
#define MOD MM+7
#define pi pair<int32,int32>
#define pl pair<int64,int64>
#define chmax(a,b) (a<b ? a=b:0)
#define chmin(a,b) (a>b ? a=b:0)
const long double PI = acos(-1);
const long long INF = 1e15;
int dx[8] = {-1,0,1,0,-1,-1,1,1};
int dy[8] = {0,-1,0,1,-1,1,1,-1};
template<typename T>
T GCD(T u,T v){
    return v ? GCD(v,u%v) : u;
}
template<typename T>
T LCM(T x,T y){
    return x*y/(T)GCD(x,y);
}
int main(){
    long long n; cin >> n;
    long long a[n+1],b[n+1];
    fill(a,a+n+1,0);
    fill(b,b+n+1,0);
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    int m = 0;
    vector<int> v(0);
    for(int i = n; i >= 2; i--){
        int num = 0;
        for(int j = i; j <= n; j += i){
            num += b[j];
        }
        if (a[i] != num % 2){
            v.push_back(i);
            b[i] = 1;
        } else {
            b[i] = 0;
        }
    }
    int c = 0;
    for(int i = 2; i <= n; i++){
        c += b[i];
    }
    if (c % 2 != a[1]){
        v.push_back(1);
    }
    cout << v.size() << endl;
    for(int i = 0; i < v.size(); i++){
        cout << v[i] << " ";
    }
    cout << endl;
}