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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <numeric>
#include <string>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <bitset>
#include <iomanip> 

using ll = long long;
using ull = unsigned long long;
using namespace std;

#define FALSE printf("false\n");
#define TRUE printf("true\n");
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i = 0;(i) < ((int)(n));++(i))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() )
#define test(a) cout<<"line:"<<__LINE__<<"["<<(#a)<<": "<<(a)<<"]"<<endl

const int INF = 1e9+7;
const ll INFL = 9*1e18;

// const int dx[4] = { 1, 0, -1, 0 };
// const int dy[4] = { 0, 1, 0, -1 };

const int dx[8] = {1, 1, 0,-1,-1,-1, 0, 1};
const int dy[8] = {0, 1, 1, 1, 0,-1,-1,-1};

ll inline digit(ll num){int tmp = 0;while(num){tmp++;num/=10;}return tmp;} // 桁数
template<typename T>inline void print(T&& x){cout<<x<<endl;}// 改行付き出力
template<typename T>inline T SUM(vector<T> vec){return accumulate(all(vec),(T)0);} // vectorの中身を全部足す
template<typename T>inline T digitSum(T num){T sum = 0;while(num){sum+=num%10;num/=10;}return sum;} // 各桁の和
template<typename T>inline T gcd(T a,T b){if(b == 0)return a;return gcd(b,a%b);} // 最大公約数
template<typename T>inline T lcm(T a, T b){T g = gcd(a,b);return a/g*b;} // 最小公倍数
template<typename T>inline T power(T a,T b){T tmp=1;rep(i,b){tmp*=a;}return tmp;} // 累乗
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
int ctoi(char c){return c-'0';}


vector<ll> primes(ll n){
  vector<ll> v;
  for (ll i = 2; i*i <= n; ++i) {
    while(n%i==0){
      n/=i;
      v.pb(i);
    }
  }
  if(n != 1)v.pb(n);
  return v;
}

// TODO: write
int main() {
  int n;cin>>n;
  vector<int> a(n);rep(i,n)cin>>a[i];
  vector<int> b(n,0);
  
  for(int i = n-1;i >= 0;--i){
    if(b[i]%2 == a[i])continue;
    auto p = primes(i);
    for(auto&& one: p){
      b[one]++;
    }
  }

  vector<int> ans;
  rep(i,n){
    if(b[i]%2){
      ans.pb(i+1);
    }
  }

  print(ans.size());
  for(auto&& one: ans){
    cout<<one<<" ";
  }
  cout<<endl;

  return 0;
}