#include <bits/stdc++.h>
using namespace std;
template <class T>
void mini(T& a, T b) {
  a = min(a, b);
}
template <class T>
void maxi(T& a, T b) {
  a = max(a, b);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  return out << a.first << " " << a.second;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (auto& i : a) out << i << " ";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, set<T> a) {
  for (auto& i : a) out << i << " ";
  return out;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, map<T1, T2> a) {
  for (auto& i : a) out << i << " ";
  return out;
}
template <class T>
void out(T a) {
  cout << a << endl;
}
void out() { cout << "OK" << endl; }
const int N = 1e5 + 5;
vector<int> g[N];
int a[N];
vector<int> ans;
vector<int> path;
int ou[N];
int mx;
bool dfs(int v, int tp = 0) {
  path.push_back(v);
  if (!tp) maxi(mx, (int)path.size());
  if (tp) {
    if (path.size() == mx) {
      ans = path;
      return 1;
    }
  }
  for (int u : g[v]) {
    if (a[u] || ou[u] > 1) continue;
    if (dfs(u, tp)) return 1;
  }
  path.pop_back();
  return 0;
}
void Solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x) {
      g[i].push_back(x);
      ou[x]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i]) {
      dfs(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i]) {
      if (dfs(i, 1)) break;
    }
  }
  reverse(ans.begin(), ans.end());
  out(ans.size());
  out(ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  Solve();
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
template <class T>
void mini(T& a, T b) {
  a = min(a, b);
}
template <class T>
void maxi(T& a, T b) {
  a = max(a, b);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  return out << a.first << " " << a.second;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (auto& i : a) out << i << " ";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, set<T> a) {
  for (auto& i : a) out << i << " ";
  return out;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, map<T1, T2> a) {
  for (auto& i : a) out << i << " ";
  return out;
}
template <class T>
void out(T a) {
  cout << a << endl;
}
void out() { cout << "OK" << endl; }
const int N = 1e5 + 5;
vector<int> g[N];
int a[N];
vector<int> ans;
vector<int> path;
int ou[N];
int mx;
bool dfs(int v, int tp = 0) {
  path.push_back(v);
  if (!tp) maxi(mx, (int)path.size());
  if (tp) {
    if (path.size() == mx) {
      ans = path;
      return 1;
    }
  }
  for (int u : g[v]) {
    if (a[u] || ou[u] > 1) continue;
    if (dfs(u, tp)) return 1;
  }
  path.pop_back();
  return 0;
}
void Solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x) {
      g[i].push_back(x);
      ou[x]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i]) {
      dfs(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i]) {
      if (dfs(i, 1)) break;
    }
  }
  reverse(ans.begin(), ans.end());
  out(ans.size());
  out(ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  Solve();
  return 0;
}

```