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 <iostream>
#include <vector>
#include <iomanip>
#include <string>
#include <sstream>
#include <algorithm>
#include <map>
#include <bits/stdc++.h>

using namespace std;
#define REP(i,n) FOR(i,0,n)
#define FOR(i,a,b) for(long long i=(a),i##Len_=(b);i<i##Len_;i++)
typedef long long ll;
static const ll MOD = 1000000007;
static const ll INF = 1000000000000000000LL;

int main() {

    ll N;
    cin >> N;

    ll a[N + 1];
    ll b[N + 1];//0 or 1

    for(ll i = 1; i <= N; ++i)
    {
        cin >> a[i];
    }

    for(ll i = N; i >= 1; --i)
    {
        ll tmp = 0;
        ll j = i;
        while(j <= N)
        {
            tmp += a[j];
            j = j + i;//倍数を取りに行く
        }
        b[i] = tmp % 2;

    }
    
    //1を探す
    ll cnt = 0;

    for(ll i = 1; i <= N; ++i)
    {
        if(b[i] == 1)++cnt;
    }

    cout << cnt << endl;

    if(cnt != 0)
    {
        for(ll i = 1; i <= N; ++i)
        {
            if(b[i] == 1){
                cout << i << ' ';
            }
        }
    }

    return 0;
}