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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

#define REP(i,m,n) for(int (i)=(m),i##_len=(n);(i)<(i##_len);++(i))
#define FORR(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define PRINT(x) cout << (x) << endl
#define SZ(x) ((int)(x).size())
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))
#define pb push_back

int main()
{
    ll N;
    cin >> N;

    vector<ll> a(N+1);
    REP(i,1,N+1) cin >> a[i];

    vector<ll> b(N+1);
    ll s;
    for(ll i=N; i>=1; --i){
        s = 0;
        for(ll j=i+i; j<=N; j+=i) s += b[j];
        b[i] = (s + a[i]) % 2;
    }

    ll cnt = 0;
    REP(i,1,N+1)
        if(b[i] == 1) ++cnt;
    cout << cnt << endl;
    REP(i,1,N+1)
        if(b[i] == 1) cout << i;
    cout << endl;

    return 0;
}