#include <bits/stdc++.h>
using namespace std;
long long vis[100500];
long long t[100500];
long long a[100500];
long long par[100500];
long long val[100500];
vector<long long> to[100500];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fill(par, par + 100500, -1);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> t[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (!a[i]) continue;
    a[i]--;
    par[i] = a[i];
    to[a[i]].push_back(i);
  }
  for (long long i = 0; i < n; i++) {
    if (to[i].size() > 1) {
      for (auto u : to[i]) {
        par[u] = -1;
      }
    }
  }
  queue<long long> q;
  for (long long i = 0; i < n; i++) {
    if (par[i] == -1) {
      q.push(i);
      val[i] = 1;
      continue;
    }
  }
  long long ans = -1, res;
  while (!q.empty()) {
    long long tmp = q.front();
    vis[tmp] = 1;
    q.pop();
    if (to[tmp].size() > 1) {
      for (auto u : to[tmp]) {
        if (vis[u]) continue;
        val[u] = 1;
        q.push(u);
        par[u] = -1;
        vis[u] = 1;
      }
    } else if (to[tmp].size() == 1) {
      if (vis[to[tmp][0]]) continue;
      val[to[tmp][0]] = val[tmp] + 1;
      q.push(to[tmp][0]);
    } else {
      if (t[tmp] == 1) {
        if (val[tmp] > ans) {
          ans = val[tmp];
          res = tmp;
        }
      }
    }
  }
  assert(ans != -1);
  vector<long long> v;
  while (res != -1) {
    v.push_back(res + 1);
    res = par[res];
  }
  reverse(v.begin(), v.end());
  cout << v.size() << '\n';
  for (auto now : v) {
    cout << now << ' ';
  }
  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;
long long vis[100500];
long long t[100500];
long long a[100500];
long long par[100500];
long long val[100500];
vector<long long> to[100500];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fill(par, par + 100500, -1);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> t[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (!a[i]) continue;
    a[i]--;
    par[i] = a[i];
    to[a[i]].push_back(i);
  }
  for (long long i = 0; i < n; i++) {
    if (to[i].size() > 1) {
      for (auto u : to[i]) {
        par[u] = -1;
      }
    }
  }
  queue<long long> q;
  for (long long i = 0; i < n; i++) {
    if (par[i] == -1) {
      q.push(i);
      val[i] = 1;
      continue;
    }
  }
  long long ans = -1, res;
  while (!q.empty()) {
    long long tmp = q.front();
    vis[tmp] = 1;
    q.pop();
    if (to[tmp].size() > 1) {
      for (auto u : to[tmp]) {
        if (vis[u]) continue;
        val[u] = 1;
        q.push(u);
        par[u] = -1;
        vis[u] = 1;
      }
    } else if (to[tmp].size() == 1) {
      if (vis[to[tmp][0]]) continue;
      val[to[tmp][0]] = val[tmp] + 1;
      q.push(to[tmp][0]);
    } else {
      if (t[tmp] == 1) {
        if (val[tmp] > ans) {
          ans = val[tmp];
          res = tmp;
        }
      }
    }
  }
  assert(ans != -1);
  vector<long long> v;
  while (res != -1) {
    v.push_back(res + 1);
    res = par[res];
  }
  reverse(v.begin(), v.end());
  cout << v.size() << '\n';
  for (auto now : v) {
    cout << now << ' ';
  }
  return 0;
}

```