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;
const double pi = acos(-1);
#define FOR(i,a,b) for (ll i=(a),__last_##i=(b);i<__last_##i;i++)
#define RFOR(i,a,b) for (ll i=(b)-1,__last_##i=(a);i>=__last_##i;i--)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define __GET_MACRO3(_1, _2, _3, NAME, ...) NAME
#define rep(...) __GET_MACRO3(__VA_ARGS__, FOR, REP)(__VA_ARGS__)
#define rrep(...) __GET_MACRO3(__VA_ARGS__, RFOR, RREP)(__VA_ARGS__)
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v) {
REP(i,v.size()){if(i)os<<" ";os<<v[i];}return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v) {
REP(i,v.size()){if(i)os<<endl;os<<v[i];}return os;}

int main() {
	int n; cin >> n;
	std::vector<int> a;
	for (int i = 0; i < n; ++i)
	{
		int an; cin >> an;
		a.push_back(an);
	}
	int b[n] = {999};


	

	for (int i = 1; i < n+1; ++i)
	{
		for (int j = n/(i+1); j < n/i; ++j)
		{
			int sum;
			int k = 1;
			while( (j+1)*k <= n ) {
				sum += a[ (j+1)*k - 1];
				k++;
			}
			sum = sum - a[j];
			
			if(a[j] == sum%2){
				b[j] = 0;
			}else{
				b[j] = 1;
			}
		}
	}

	int count = 0;
	std::vector<int> ans;
	for (int i = 0; i < n; ++i)
	{
		if (b[i] == 1)
		{
			ans.push_back(i+1);
			count++;
		}
	}

	cout << count << endl;
	if(count != 0) {
		
		for (int i = 0; i < (int)ans.size() - 1; ++i)
		{
			cout << ans[i] << ' ';
		}
		cout << ans[(int)ans.size() - 1] << endl;
	}
	

	











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