#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<int> ishotel(n);
  vector<int> isvisited(n, 0);
  vector<bool> dead(n, false);
  vector<int> kill(n, 0);
  vector<int> child(n);
  vector<int> len(n, 0);
  vector<int> parent(n, -1);
  vector<vector<int> > g(n);
  queue<int> oq, q;
  for (int i = 0; i < n; i++) cin >> ishotel[i];
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    x--;
    if (x != -1) {
      g[i].push_back(x);
      kill[x]++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (kill[i] != 1) {
      dead[i] = true;
      isvisited[i] = 1;
    }
    if (ishotel[i] == 1) {
      child[i] = -1;
      oq.push(i);
    }
  }
  int x = 0;
  vector<int> ans;
  while (!oq.empty()) {
    q.push(oq.front());
    len[oq.front()] = 1;
    oq.pop();
    int turn = 0;
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      turn++;
      for (int i = 0; i < g[v].size(); i++) {
        int f = g[v][i];
        if (!dead[f]) {
          q.push(f);
          parent[f] = v;
          len[f] = max(len[f], len[v] + 1);
        }
      }
    }
  }
  int maxind = 0;
  for (int i = 1; i < n; i++)
    if (len[i] > len[maxind]) maxind = i;
  printf("%d\n", len[maxind]);
  int j = maxind;
  do {
    ans.push_back(j);
    j = parent[j];
  } while (j != -1);
  for (int i = 0; i < ans.size(); i++)
    printf("%d%c", ans[i] + 1, ((i == ans.size() - 1) ? '\n' : ' '));
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
int main() {
  int n;
  cin >> n;
  vector<int> ishotel(n);
  vector<int> isvisited(n, 0);
  vector<bool> dead(n, false);
  vector<int> kill(n, 0);
  vector<int> child(n);
  vector<int> len(n, 0);
  vector<int> parent(n, -1);
  vector<vector<int> > g(n);
  queue<int> oq, q;
  for (int i = 0; i < n; i++) cin >> ishotel[i];
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    x--;
    if (x != -1) {
      g[i].push_back(x);
      kill[x]++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (kill[i] != 1) {
      dead[i] = true;
      isvisited[i] = 1;
    }
    if (ishotel[i] == 1) {
      child[i] = -1;
      oq.push(i);
    }
  }
  int x = 0;
  vector<int> ans;
  while (!oq.empty()) {
    q.push(oq.front());
    len[oq.front()] = 1;
    oq.pop();
    int turn = 0;
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      turn++;
      for (int i = 0; i < g[v].size(); i++) {
        int f = g[v][i];
        if (!dead[f]) {
          q.push(f);
          parent[f] = v;
          len[f] = max(len[f], len[v] + 1);
        }
      }
    }
  }
  int maxind = 0;
  for (int i = 1; i < n; i++)
    if (len[i] > len[maxind]) maxind = i;
  printf("%d\n", len[maxind]);
  int j = maxind;
  do {
    ans.push_back(j);
    j = parent[j];
  } while (j != -1);
  for (int i = 0; i < ans.size(); i++)
    printf("%d%c", ans[i] + 1, ((i == ans.size() - 1) ? '\n' : ' '));
  return 0;
}

```