#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 7;
vector<int> in[MAX], out[MAX];
bool mark[MAX];
int dist[MAX];
vector<int> t;
set<int> s;
int n;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int type;
    cin >> type;
    if (type) {
      t.push_back(i);
      s.insert(i);
    }
  }
  for (int i = 1; i <= n; ++i) {
    int u;
    cin >> u;
    out[u].push_back(i);
    in[i].push_back(u);
  }
  memset(dist, -1, sizeof(dist));
  int size = t.size();
  for (int i = 0; i < size; ++i) {
    int u = t[i];
    memset(mark, false, sizeof(mark));
    queue<int> q;
    mark[u] = true;
    mark[0] = true;
    dist[u] = 0;
    q.push(u);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      int size2 = in[u].size();
      for (int j = 0; j < size2; ++j) {
        int v = in[u][j];
        if (!mark[v] && (int)out[v].size() < 2) {
          mark[v] = true;
          dist[v] = max(dist[v], dist[u] + 1);
          q.push(v);
        }
      }
    }
  }
  int longest = -1;
  for (int i = 1; i <= n; ++i) {
    longest = max(longest, dist[i]);
  }
  vector<int> res;
  for (int i = 1; i <= n; ++i) {
    if (dist[i] == longest) {
      res.push_back(i);
      while (s.count(i) == 0) {
        i = out[i][0];
        res.push_back(i);
      }
      break;
    }
  }
  cout << longest + 1 << "\n";
  int size3 = res.size();
  for (int i = 0; i < size3; ++i) {
    cout << res[i] << " ";
  }
  cout << "\n";
  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 MAX = 1e5 + 7;
vector<int> in[MAX], out[MAX];
bool mark[MAX];
int dist[MAX];
vector<int> t;
set<int> s;
int n;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int type;
    cin >> type;
    if (type) {
      t.push_back(i);
      s.insert(i);
    }
  }
  for (int i = 1; i <= n; ++i) {
    int u;
    cin >> u;
    out[u].push_back(i);
    in[i].push_back(u);
  }
  memset(dist, -1, sizeof(dist));
  int size = t.size();
  for (int i = 0; i < size; ++i) {
    int u = t[i];
    memset(mark, false, sizeof(mark));
    queue<int> q;
    mark[u] = true;
    mark[0] = true;
    dist[u] = 0;
    q.push(u);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      int size2 = in[u].size();
      for (int j = 0; j < size2; ++j) {
        int v = in[u][j];
        if (!mark[v] && (int)out[v].size() < 2) {
          mark[v] = true;
          dist[v] = max(dist[v], dist[u] + 1);
          q.push(v);
        }
      }
    }
  }
  int longest = -1;
  for (int i = 1; i <= n; ++i) {
    longest = max(longest, dist[i]);
  }
  vector<int> res;
  for (int i = 1; i <= n; ++i) {
    if (dist[i] == longest) {
      res.push_back(i);
      while (s.count(i) == 0) {
        i = out[i][0];
        res.push_back(i);
      }
      break;
    }
  }
  cout << longest + 1 << "\n";
  int size3 = res.size();
  for (int i = 0; i < size3; ++i) {
    cout << res[i] << " ";
  }
  cout << "\n";
  return 0;
}

```