#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-11;
const int INFINITE = 0x3f3f3f3f;
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline T lowbit(T n) {
  return (n ^ (n - 1)) & n;
}
template <class T>
inline int countbit(T n) {
  return (n == 0) ? 0 : (1 + countbit(n & (n - 1)));
}
typedef vector<int> VI;
typedef vector<VI> VII;
typedef vector<string> VS;
const int MAXN = 100001;
int father[MAXN], deg[MAXN], type[MAXN];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = (int)(0); i < (int)(n); i++) cin >> type[i];
  for (int i = (int)(0); i < (int)(n); i++) {
    cin >> father[i];
    --father[i];
  }
  for (int i = (int)(0); i < (int)(n); i++)
    if (father[i] != -1) {
      deg[father[i]]++;
    }
  VI res;
  res.clear();
  for (int i = (int)(0); i < (int)(n); i++)
    if (type[i]) {
      int cur = i;
      VI temp;
      temp.clear();
      do {
        temp.push_back(cur + 1);
        cur = father[cur];
      } while (cur != -1 && deg[cur] == 1);
      if (temp.size() > res.size()) res = temp;
    }
  reverse((res).begin(), (res).end());
  cout << res.size() << endl;
  for (int i = (int)(0); i < (int)(res.size()); i++)
    cout << res[i] << (i == res.size() - 1 ? "\n" : " ");
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

Valera's finally decided to go on holiday! He packed up and headed for a ski resort.

Valera's fancied a ski trip but he soon realized that he could get lost in this new place. Somebody gave him a useful hint: the resort has n objects (we will consider the objects indexed in some way by integers from 1 to n), each object is either a hotel or a mountain.

Valera has also found out that the ski resort had multiple ski tracks. Specifically, for each object v, the resort has at most one object u, such that there is a ski track built from object u to object v. We also know that no hotel has got a ski track leading from the hotel to some object.

Valera is afraid of getting lost on the resort. So he wants you to come up with a path he would walk along. The path must consist of objects v1, v2, ..., vk (k ≥ 1) and meet the following conditions:

  1. Objects with numbers v1, v2, ..., vk - 1 are mountains and the object with number vk is the hotel. 
  2. For any integer i (1 ≤ i < k), there is exactly one ski track leading from object vi. This track goes to object vi + 1. 
  3. The path contains as many objects as possible (k is maximal). 



Help Valera. Find such path that meets all the criteria of our hero!

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of objects.

The second line contains n space-separated integers type1, type2, ..., typen — the types of the objects. If typei equals zero, then the i-th object is the mountain. If typei equals one, then the i-th object is the hotel. It is guaranteed that at least one object is a hotel.

The third line of the input contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ n) — the description of the ski tracks. If number ai equals zero, then there is no such object v, that has a ski track built from v to i. If number ai doesn't equal zero, that means that there is a track built from object ai to object i.

Output

In the first line print k — the maximum possible path length for Valera. In the second line print k integers v1, v2, ..., vk — the path. If there are multiple solutions, you can print any of them.

Examples

Input

5
0 0 0 0 1
0 1 2 3 4


Output

5
1 2 3 4 5


Input

5
0 0 1 0 1
0 1 2 2 4


Output

2
4 5


Input

4
1 0 0 0
2 3 4 2


Output

1
1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-11;
const int INFINITE = 0x3f3f3f3f;
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline T lowbit(T n) {
  return (n ^ (n - 1)) & n;
}
template <class T>
inline int countbit(T n) {
  return (n == 0) ? 0 : (1 + countbit(n & (n - 1)));
}
typedef vector<int> VI;
typedef vector<VI> VII;
typedef vector<string> VS;
const int MAXN = 100001;
int father[MAXN], deg[MAXN], type[MAXN];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = (int)(0); i < (int)(n); i++) cin >> type[i];
  for (int i = (int)(0); i < (int)(n); i++) {
    cin >> father[i];
    --father[i];
  }
  for (int i = (int)(0); i < (int)(n); i++)
    if (father[i] != -1) {
      deg[father[i]]++;
    }
  VI res;
  res.clear();
  for (int i = (int)(0); i < (int)(n); i++)
    if (type[i]) {
      int cur = i;
      VI temp;
      temp.clear();
      do {
        temp.push_back(cur + 1);
        cur = father[cur];
      } while (cur != -1 && deg[cur] == 1);
      if (temp.size() > res.size()) res = temp;
    }
  reverse((res).begin(), (res).end());
  cout << res.size() << endl;
  for (int i = (int)(0); i < (int)(res.size()); i++)
    cout << res[i] << (i == res.size() - 1 ? "\n" : " ");
  return 0;
}

```