#include <bits/stdc++.h>
using namespace std;
vector<long long int> ans;
pair<long long int, long long int> dfs(
    vector<vector<long long int> > &graph, long long int s, vector<bool> &vis,
    vector<long long int> &type,
    vector<pair<long long int, long long int> > &dp) {
  if (vis[s]) {
    return dp[s];
  }
  vis[s] = true;
  if (graph[s].size() == 0) {
    if (type[s] == 1) {
      dp[s].first = s;
      dp[s].second = 1;
      return make_pair(s, 1);
    }
    dp[s].first = -1;
    dp[s].second = -1;
    return make_pair(-1, -1);
  }
  if (type[s] == 1) {
    dp[s].first = s;
    dp[s].second = 1;
    return make_pair(s, 1);
  }
  if (graph[s].size() > 1) {
    dp[s].first = -1;
    dp[s].second = -1;
    return make_pair(-1, -1);
  }
  pair<long long int, long long int> p = dfs(graph, graph[s][0], vis, type, dp);
  if (p.first == -1) {
    dp[s].first = -1;
    dp[s].second = -1;
    return make_pair(-1, -1);
  }
  dp[s].first = p.first;
  dp[s].second = p.second + 1;
  return dp[s];
}
void dfs(vector<vector<long long int> > &graph, long long int s,
         long long int f) {
  ans.push_back(s);
  if (s == f) {
    return;
  }
  return dfs(graph, graph[s][0], f);
}
void code() {
  long long int n;
  cin >> n;
  vector<long long int> type(n + 1);
  for (long long int i = 1; i <= n; i++) cin >> type[i];
  vector<vector<long long int> > graph(n + 1);
  for (long long int i = 1; i <= n; i++) {
    long long int u;
    cin >> u;
    if (u != 0) {
      graph[u].push_back(i);
    }
  }
  long long int len = 0, s = 0, e = 0;
  vector<bool> vis(n + 1, false);
  vector<pair<long long int, long long int> > dp(n + 1);
  for (long long int i = 1; i <= n; i++) {
    dp[i].first = -1;
    dp[i].second = -1;
  }
  for (long long int i = 1; i <= n; i++) {
    if (!vis[i]) {
      pair<long long int, long long int> p = dfs(graph, i, vis, type, dp);
    }
  }
  for (long long int i = 1; i <= n; i++) {
    if (dp[i].second > len) {
      len = dp[i].second;
      s = i;
      e = dp[i].first;
    }
  }
  dfs(graph, s, e);
  cout << len << "\n";
  for (auto i : ans) cout << i << " ";
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int tc = 1;
  while (tc--) {
    code();
  }
  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;
vector<long long int> ans;
pair<long long int, long long int> dfs(
    vector<vector<long long int> > &graph, long long int s, vector<bool> &vis,
    vector<long long int> &type,
    vector<pair<long long int, long long int> > &dp) {
  if (vis[s]) {
    return dp[s];
  }
  vis[s] = true;
  if (graph[s].size() == 0) {
    if (type[s] == 1) {
      dp[s].first = s;
      dp[s].second = 1;
      return make_pair(s, 1);
    }
    dp[s].first = -1;
    dp[s].second = -1;
    return make_pair(-1, -1);
  }
  if (type[s] == 1) {
    dp[s].first = s;
    dp[s].second = 1;
    return make_pair(s, 1);
  }
  if (graph[s].size() > 1) {
    dp[s].first = -1;
    dp[s].second = -1;
    return make_pair(-1, -1);
  }
  pair<long long int, long long int> p = dfs(graph, graph[s][0], vis, type, dp);
  if (p.first == -1) {
    dp[s].first = -1;
    dp[s].second = -1;
    return make_pair(-1, -1);
  }
  dp[s].first = p.first;
  dp[s].second = p.second + 1;
  return dp[s];
}
void dfs(vector<vector<long long int> > &graph, long long int s,
         long long int f) {
  ans.push_back(s);
  if (s == f) {
    return;
  }
  return dfs(graph, graph[s][0], f);
}
void code() {
  long long int n;
  cin >> n;
  vector<long long int> type(n + 1);
  for (long long int i = 1; i <= n; i++) cin >> type[i];
  vector<vector<long long int> > graph(n + 1);
  for (long long int i = 1; i <= n; i++) {
    long long int u;
    cin >> u;
    if (u != 0) {
      graph[u].push_back(i);
    }
  }
  long long int len = 0, s = 0, e = 0;
  vector<bool> vis(n + 1, false);
  vector<pair<long long int, long long int> > dp(n + 1);
  for (long long int i = 1; i <= n; i++) {
    dp[i].first = -1;
    dp[i].second = -1;
  }
  for (long long int i = 1; i <= n; i++) {
    if (!vis[i]) {
      pair<long long int, long long int> p = dfs(graph, i, vis, type, dp);
    }
  }
  for (long long int i = 1; i <= n; i++) {
    if (dp[i].second > len) {
      len = dp[i].second;
      s = i;
      e = dp[i].first;
    }
  }
  dfs(graph, s, e);
  cout << len << "\n";
  for (auto i : ans) cout << i << " ";
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int tc = 1;
  while (tc--) {
    code();
  }
  return 0;
}

```