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>
#include <type_traits>
using namespace std;

using ll = int64_t;
#define int ll

#define FOR(i, a, b) for (int i = int(a); i < int(b); i++)
#define REP(i, b) FOR(i, 0, b)
#define MP make_pair
#define PB push_back
#define EB emplace_back
#define ALL(x) x.begin(), x.end()

using pi = pair<int, int>;
using vi = vector<int>;
using ld = long double;

template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}

template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  REP(i, (int)v.size()) {
    if (i) os << ",";
    os << v[i];
  }
  os << "}";
  return os;
}

ll read() {
  ll i;
  scanf("%" SCNd64, &i);
  return i;
}

void printSpace() { printf(" "); }

void printEoln() { printf("\n"); }

void print(ll x, int suc = 1) {
  printf("%" PRId64, x);
  if (suc == 1) printEoln();
  if (suc == 2) printSpace();
}

string readString() {
  static char buf[3341000];
  scanf("%s", buf);
  return string(buf);
}

char *readCharArray() {
  static char buf[3341000];
  static int bufUsed = 0;
  char *ret = buf + bufUsed;
  scanf("%s", ret);
  bufUsed += strlen(ret) + 1;
  return ret;
}

template <class T, class U>
void chmax(T &a, U b) {
  if (a < b) a = b;
}

template <class T, class U>
void chmin(T &a, U b) {
  if (b < a) a = b;
}

template <class T>
T Sq(const T &t) {
  return t * t;
}

const int mod = 1e9 + 7;

signed main() {
  int n = read();
  vi a(n + 1), b(n + 1);
  REP(i, n) { a[i + 1] = read(); }
  int m = 0;
  for (int i = n; i > 0; i--) {
    int j = i + i;
    if (j > n) {
      b[i] = a[i];
    } else {
      while (j <= n) {
        b[i] += b[j];
        j += i;
      }
      b[i] &= 1;
      if (b[i] == a[i])
        b[i] = 0;
      else
        b[i] = 1;
    }
    m += b[i];
  }
  print(n);
  FOR(i, 1, n + 1) {
    if (b[i] == 1) print(i);
  }
}