#include <bits/stdc++.h>
using namespace std;
vector<int> v;
vector<int> v_f;
int N;
int t[100005];
int fin[100005];
int odeg[100005];
bool vis[100005];
int func(int a) {
  if (odeg[a] > 1) return 0;
  if (vis[a]) return 0;
  if (fin[a] == 0) {
    vis[a] = 1;
    v.push_back(a);
    return 1;
  }
  vis[a] = 1;
  v.push_back(a);
  return 1 + func(fin[a]);
}
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> t[i];
  for (int i = 1; i <= N; i++) {
    cin >> fin[i];
    odeg[fin[i]] += 1;
  }
  int final = 0;
  int cur;
  for (int i = 1; i <= N; i++) {
    if (t[i]) {
      v.clear();
      cur = func(i);
      if (cur > final) {
        final = cur;
        v_f.clear();
        v_f = vector<int>(v.begin(), v.end());
      }
    }
  }
  cout << final << endl;
  cout << v_f[final - 1];
  for (int i = final - 2; i >= 0; i--) {
    cout << " " << v_f[i];
  }
  cout << endl;
}

### Prompt

Generate 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;
vector<int> v;
vector<int> v_f;
int N;
int t[100005];
int fin[100005];
int odeg[100005];
bool vis[100005];
int func(int a) {
  if (odeg[a] > 1) return 0;
  if (vis[a]) return 0;
  if (fin[a] == 0) {
    vis[a] = 1;
    v.push_back(a);
    return 1;
  }
  vis[a] = 1;
  v.push_back(a);
  return 1 + func(fin[a]);
}
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> t[i];
  for (int i = 1; i <= N; i++) {
    cin >> fin[i];
    odeg[fin[i]] += 1;
  }
  int final = 0;
  int cur;
  for (int i = 1; i <= N; i++) {
    if (t[i]) {
      v.clear();
      cur = func(i);
      if (cur > final) {
        final = cur;
        v_f.clear();
        v_f = vector<int>(v.begin(), v.end());
      }
    }
  }
  cout << final << endl;
  cout << v_f[final - 1];
  for (int i = final - 2; i >= 0; i--) {
    cout << " " << v_f[i];
  }
  cout << endl;
}

```