#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int MAX = 100010;
int n;
bool type[MAX];
int outDegree[MAX], parent[MAX];
int main() {
  cin >> n;
  for (int i = 0, _n = (n); i < _n; ++i) {
    cin >> type[i];
  }
  for (int i = 0, _n = (n); i < _n; ++i) {
    cin >> parent[i];
    parent[i]--;
    if (parent[i] != -1) outDegree[parent[i]]++;
  }
  vector<int> ans;
  for (int i = 0, _n = (n); i < _n; ++i) {
    if (type[i] == 1) {
      int curV = i;
      vector<int> cur;
      while (outDegree[parent[curV]] <= 1 && parent[curV] != -1) {
        cur.push_back(curV);
        curV = parent[curV];
      }
      cur.push_back(curV);
      if (ans.size() < cur.size()) ans = cur;
    }
  }
  cout << (int)(ans).size() << endl;
  reverse((ans).begin(), (ans).end());
  for (int i = 0, _n = ((int)(ans).size()); i < _n; ++i) {
    cout << ans[i] + 1 << " ";
  }
  cout << endl;
  return 0;
}

### 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;
const int MOD = 1e9 + 7;
const int MAX = 100010;
int n;
bool type[MAX];
int outDegree[MAX], parent[MAX];
int main() {
  cin >> n;
  for (int i = 0, _n = (n); i < _n; ++i) {
    cin >> type[i];
  }
  for (int i = 0, _n = (n); i < _n; ++i) {
    cin >> parent[i];
    parent[i]--;
    if (parent[i] != -1) outDegree[parent[i]]++;
  }
  vector<int> ans;
  for (int i = 0, _n = (n); i < _n; ++i) {
    if (type[i] == 1) {
      int curV = i;
      vector<int> cur;
      while (outDegree[parent[curV]] <= 1 && parent[curV] != -1) {
        cur.push_back(curV);
        curV = parent[curV];
      }
      cur.push_back(curV);
      if (ans.size() < cur.size()) ans = cur;
    }
  }
  cout << (int)(ans).size() << endl;
  reverse((ans).begin(), (ans).end());
  for (int i = 0, _n = ((int)(ans).size()); i < _n; ++i) {
    cout << ans[i] + 1 << " ";
  }
  cout << endl;
  return 0;
}

```