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;

#define rep(i,n) for(ll i = 0; i < n; i++)
#define REP(i,n) for(ll i = 1; i < n; i++)
#define sz(x) ll(x.size())
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define YesorNo(a) printf(a ? "Yes\n" : "No\n")
#define endl '\n'
using ll = long long;
using P = pair<ll,ll>;
using Pl = pair<ll,ll>;
using vi = vector<ll>;
using vii = vector<vi>;
using vl = vector<ll>;
using vll = vector<vl>;
using vs = vector<string>;
using vb = vector<bool>;
using vc = vector<char>;
using vcc = vector<vc>;
const ll dx[] = {0,1,0,-1,1,1,-1,-1};
const ll dy[] = {1,0,-1,0,1,-1,-1,1};
const ll inf = (1<<30)-1;
const ll infll = (1LL<<62)-1;
ll ceil(const ll &a, const ll &b){return ((a)+(b)-1)/b;}
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

bool isprime(ll x){
  ll i;
  if(x < 2) return false;
  else if(x == 2) return true;
  if(x % 2 == 0) return false;
  for(i = 3; i*i <= x; i += 2){
    if(x % i == 0) return false;
  }
  return true;
}

int main(){
  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  ll n;
  cin >> n;
  vi a(n);
  rep(i,n) cin >> a[i];
  vi prime;
  REP(i,n+1) {
    if(isprime(i)) prime.push_back(i);
  }
  vi ans(n);
  ll m = 0;
  for(ll i = n-1; i >= 0; i--) {
    ll sum = 0;
    for(ll p : prime) {
      if((i+1)*p > n) break;
      sum += a[(i+1)*p-1];
    }
    sum %= 2;
    if(sum != a[i]) {
      m++;
      ans[i]++;
    }
  }
  cout << m << endl;
  rep(i,n) {
    if(ans[i]) cout << i+1 << ' ';
  }
}