#include <bits/stdc++.h>
using namespace std;
unordered_set<long long int> adj[100005], adj1[100005];
vector<bool> vis(100005, false);
vector<long long int> dist(100005, -1), en(100005, 0);
long long int k;
vector<long long int> par(100005, -1);
void bfs(long long int s, bool flag = false) {
  queue<long long int> q;
  q.push(s);
  vis[s] = true;
  dist[s] = 1;
  par[s] = -1;
  while (!q.empty()) {
    s = q.front();
    if (flag) {
      cout << s << " ";
      if (dist[s] == k) return;
    }
    q.pop();
    for (auto ch : adj[s]) {
      if (dist[ch] == -1 && en[ch] == 1) {
        dist[ch] = 1 + dist[s];
        q.push(ch);
        vis[ch] = true;
        par[ch] = s;
      }
    }
  }
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> tp(n + 1);
  for (long long int i = 1; i < n + 1; i++) cin >> tp[i];
  for (long long int i = 1; i < n + 1; i++) {
    long long int u;
    cin >> u;
    if (u == 0) continue;
    adj[i].insert(u);
    en[u]++;
  }
  for (long long int i = 1; i < n + 1; i++) {
    if (!vis[i] && tp[i] == 1) bfs(i);
  }
  k = 0;
  long long int src;
  for (long long int i = 1; i < n + 1; i++) {
    if (k < dist[i]) {
      k = dist[i];
      src = i;
    }
  }
  vis.assign(vis.size(), false);
  dist.assign(dist.size(), -1);
  cout << k << "\n";
  while (src != -1) {
    cout << src << " ";
    src = par[src];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
unordered_set<long long int> adj[100005], adj1[100005];
vector<bool> vis(100005, false);
vector<long long int> dist(100005, -1), en(100005, 0);
long long int k;
vector<long long int> par(100005, -1);
void bfs(long long int s, bool flag = false) {
  queue<long long int> q;
  q.push(s);
  vis[s] = true;
  dist[s] = 1;
  par[s] = -1;
  while (!q.empty()) {
    s = q.front();
    if (flag) {
      cout << s << " ";
      if (dist[s] == k) return;
    }
    q.pop();
    for (auto ch : adj[s]) {
      if (dist[ch] == -1 && en[ch] == 1) {
        dist[ch] = 1 + dist[s];
        q.push(ch);
        vis[ch] = true;
        par[ch] = s;
      }
    }
  }
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> tp(n + 1);
  for (long long int i = 1; i < n + 1; i++) cin >> tp[i];
  for (long long int i = 1; i < n + 1; i++) {
    long long int u;
    cin >> u;
    if (u == 0) continue;
    adj[i].insert(u);
    en[u]++;
  }
  for (long long int i = 1; i < n + 1; i++) {
    if (!vis[i] && tp[i] == 1) bfs(i);
  }
  k = 0;
  long long int src;
  for (long long int i = 1; i < n + 1; i++) {
    if (k < dist[i]) {
      k = dist[i];
      src = i;
    }
  }
  vis.assign(vis.size(), false);
  dist.assign(dist.size(), -1);
  cout << k << "\n";
  while (src != -1) {
    cout << src << " ";
    src = par[src];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```