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 <map>
#include <algorithm>
#include <fstream>
#include<cstdio>
#include<iomanip>

#define REP(i, n) for(int i=0;i<n;i++)

using namespace std;

typedef long long ll;
typedef unsigned long ul;
typedef long double ld;

const int maxN = 200000;

ll B[maxN];
int main() {
    ll N;
    cin >> N;
    ll A[N];
    REP(i, N) {
        cin >> A[i];
    }
    REP(i, N) {
        ll r = N / (N - i);//rは，「N-i番目の数の，倍数の個数（自分含む）」
        if (r == 1) {
            B[N - i - 1] = A[N - i - 1];
        } else {
            ll sum = 0;
            for (int m = 2; m < r + 1; m++) {
                sum += B[m * (N - i) - 1];
            }
            if (A[N - i - 1] != sum % 2) {
                B[N - i - 1] = 1;
            }
        }
    }

    ll M = 0;
    REP(i, N) {
        M += B[i];
    }
    cout <<M<<endl;
    REP(i,N){
        if(B[i]==1){
            cout<<i+1;
            cout<<" ";
        }
    }
}