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 <algorithm>
#include <string>
#include <vector>
#include <set>
#include <cmath>
#include <stdio.h>
#include <list>
#include <numeric>
#include <stack>
#include <queue>
#include <tuple>
#include <bitset>
#include <map>
#include <math.h>
//　ceil(a/b)   (a + (b - 1))/ b
using namespace std;
#define rep(i,n) for(int i = 0; i < (n); i++)
typedef long long ll;
typedef pair<int,int> P;
const int inf=1000000007;
const ll mod=1000000007;
const double PI=3.14159265358979323846;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};

ll gcd(ll a, ll b){
  if (b==0) return a;
  else return gcd(b,a%b);
}


// ll memo[2020][2020];
// int pascal_triangle(){
//   for(int i = 0; i < 1000; i++){
//     for(int j = 0; j < i+1; j++){
//       if(j == 0 || j == i) memo[i][j] = 1;
//       else{
// 	memo[i][j] = (memo[i-1][j-1]+memo[i-1][j])%mod;
//       }
//     }
//   }
//   return 0;
// }


// int factorial(int n){
//   if(n == 0) return 1;
//   return n*factorial(n-1);
// }

// int divisor(int a){
//   int cnt = 0;
//   for(int i=1; i <= a; i++) if(a%i == 0) cnt++;
//   return cnt;
// }

// int comb(int n, int k){
//   pascal_triangle();
//   return memo[n][k];
// }

int cans(bool f){
  if(f) cout << "Yes" << endl;
  else cout << "No" << endl;
  return 0;
}







//二項係数(mod剰余)----------------------------

// const int MAX = 3000010;
// const int M = 1000000007;

// vector<ll> fac(MAX); //n!(mod M)
// //a,bの範囲的にこれだけ配列を用意していけば十分
// vector<ll> ifac(MAX); //k!^{M-2} (mod M)

// ll mpow(ll x, ll n){ //x^n(mod M) ←普通にpow(x,n)では溢れてしまうため，随時mod計算
//     ll ans = 1;
//     while(n != 0){
//         if(n&1) ans = ans*x % M;
//         x = x*x % M;
//         n = n >> 1;
//     }
//     return ans;
// }

// ll COMinit(){
//   //大した量ではないので，先にfax[i]とifax[i]を全て計算しておく
//   fac[0] = 1;
//   ifac[0] = 1;
//   for(ll i = 0; i<MAX; i++){
//     fac[i+1] = fac[i]*(i+1) % M; // n!(mod M)
//     ifac[i+1] = ifac[i]*mpow(i+1, M-2) % M; // k!^{M-2} (mod M) ←累乗にmpowを採用
//   }
//   return 0;
// }

// ll COM(ll a, ll b){ //aCbをmod計算
//   if(a == 0 && b == 0)return 1;
//   if(a < b || a < 0)return 0;
//   ll tmp = ifac[a-b]* ifac[b] % M;
//   return tmp * fac[a] % M;
// }

//-----------------------------------------

int b[1000000];

int main(){
  int N;
  cin >> N;
  int a[N+1];
  rep(i,N) cin >> a[i+1];
  for(int i = N; i > 0; i--){
    int cnt = N/i;
    int sum = 0;
    rep(j,cnt){
      sum += b[(j+1)*i];
      sum %= 2;
    }
    if(sum == a[i]) b[i] = 0;
    else b[i] = 1;
  }
  vector<int> v;
  int ans = 0;
  rep(i,N+1) if(b[i]){
    ans++;
    v.push_back(i);
  } 
  cout << ans << endl;
  if(ans){
    rep(i,N){
      if(b[i+1]) cout << i+1 << " ";
    }
  }
  return 0;
}