#include <bits/stdc++.h>
using namespace std;
long long int n, i, j = 0, k, obj[100005], visited[100005] = {0}, a, b,
                    hotel[100005], counts = 0, key, prevs, m, temp;
vector<long long int> adj[100005], path_length[100005];
vector<long long int>::iterator it, ennd;
void dfs(long long int temp, long long int m) {
  vector<long long int>::iterator it, ennd;
  it = adj[m].begin();
  ennd = adj[m].end();
  path_length[temp].push_back(m);
  counts++;
  for (it; it != ennd; it++) {
    if (visited[*it] > 1)
      return;
    else {
      dfs(temp, *it);
    }
  }
}
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> obj[i];
    if (obj[i] == 1) hotel[++j] = i;
  }
  for (i = 1; i <= n; i++) {
    cin >> a;
    if (a == 0) continue;
    adj[i].push_back(a);
    visited[a]++;
  }
  prevs = 0;
  for (i = 1; i <= j; i++) {
    m = hotel[i];
    temp = m;
    counts = 0;
    dfs(temp, m);
    if (prevs < counts) {
      prevs = counts;
      key = temp;
    }
  }
  cout << prevs << endl;
  it = path_length[key].begin();
  ennd = path_length[key].end();
  ennd--;
  it--;
  for (ennd; ennd != it; ennd--) cout << *ennd << " ";
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long int n, i, j = 0, k, obj[100005], visited[100005] = {0}, a, b,
                    hotel[100005], counts = 0, key, prevs, m, temp;
vector<long long int> adj[100005], path_length[100005];
vector<long long int>::iterator it, ennd;
void dfs(long long int temp, long long int m) {
  vector<long long int>::iterator it, ennd;
  it = adj[m].begin();
  ennd = adj[m].end();
  path_length[temp].push_back(m);
  counts++;
  for (it; it != ennd; it++) {
    if (visited[*it] > 1)
      return;
    else {
      dfs(temp, *it);
    }
  }
}
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> obj[i];
    if (obj[i] == 1) hotel[++j] = i;
  }
  for (i = 1; i <= n; i++) {
    cin >> a;
    if (a == 0) continue;
    adj[i].push_back(a);
    visited[a]++;
  }
  prevs = 0;
  for (i = 1; i <= j; i++) {
    m = hotel[i];
    temp = m;
    counts = 0;
    dfs(temp, m);
    if (prevs < counts) {
      prevs = counts;
      key = temp;
    }
  }
  cout << prevs << endl;
  it = path_length[key].begin();
  ennd = path_length[key].end();
  ennd--;
  it--;
  for (ennd; ennd != it; ennd--) cout << *ennd << " ";
  return 0;
}

```