#include <bits/stdc++.h>
using namespace std;
long long min(long long a, int b) {
  if (a < b) return a;
  return b;
}
long long min(int a, long long b) {
  if (a < b) return a;
  return b;
}
long long max(long long a, int b) {
  if (a > b) return a;
  return b;
}
long long max(int a, long long b) {
  if (a > b) return a;
  return b;
}
const int MOD = 1e9 + 7;
const int inf = 1e9;
const long long linf = 1e18;
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
vector<int> compress(vector<int> v) {
  set<int> s;
  for (auto a : v) s.insert(a);
  vector<int> t(s.begin(), s.end());
  for (auto& a : v) a = lower_bound(t.begin(), t.end(), a) - t.begin();
  return v;
}
bool prime(long long a) {
  if (a == 1) return false;
  for (int i = 2; i * i <= a; ++i) {
    if (a % i == 0) return false;
  }
  return true;
}
long long pwr(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r *= a, r %= MOD, b--;
    a *= a, a %= MOD, b /= 2;
  }
  return r;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> ty(n), dis(n, 0), cnt(n, 0);
  vector<vector<int>> adj(n), adj2(n);
  vector<bool> vis(n);
  for (int& a : ty) cin >> a;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    if (x) --x, adj[i].push_back(x), adj2[x].push_back(i), cnt[x]++;
  }
  queue<int> q;
  for (int i = 0; i < n; ++i)
    if (ty[i]) q.push(i), vis[i] = 1, dis[i] = 1;
  int ans = 1;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    ans = max(ans, dis[v]);
    for (int u : adj[v]) {
      if (!vis[u]) {
        vis[u] = 1;
        q.push(u);
        if (cnt[u] == 1 && dis[v]) dis[u] = max(dis[u], dis[v] + 1);
      }
    }
  }
  cout << ans << '\n';
  for (int i = 0; i < n; ++i) {
    if (dis[i] == ans) {
      while (--ans) {
        cout << i + 1 << ' ';
        i = *adj2[i].begin();
      }
      cout << i + 1;
      break;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
long long min(long long a, int b) {
  if (a < b) return a;
  return b;
}
long long min(int a, long long b) {
  if (a < b) return a;
  return b;
}
long long max(long long a, int b) {
  if (a > b) return a;
  return b;
}
long long max(int a, long long b) {
  if (a > b) return a;
  return b;
}
const int MOD = 1e9 + 7;
const int inf = 1e9;
const long long linf = 1e18;
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
vector<int> compress(vector<int> v) {
  set<int> s;
  for (auto a : v) s.insert(a);
  vector<int> t(s.begin(), s.end());
  for (auto& a : v) a = lower_bound(t.begin(), t.end(), a) - t.begin();
  return v;
}
bool prime(long long a) {
  if (a == 1) return false;
  for (int i = 2; i * i <= a; ++i) {
    if (a % i == 0) return false;
  }
  return true;
}
long long pwr(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r *= a, r %= MOD, b--;
    a *= a, a %= MOD, b /= 2;
  }
  return r;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> ty(n), dis(n, 0), cnt(n, 0);
  vector<vector<int>> adj(n), adj2(n);
  vector<bool> vis(n);
  for (int& a : ty) cin >> a;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    if (x) --x, adj[i].push_back(x), adj2[x].push_back(i), cnt[x]++;
  }
  queue<int> q;
  for (int i = 0; i < n; ++i)
    if (ty[i]) q.push(i), vis[i] = 1, dis[i] = 1;
  int ans = 1;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    ans = max(ans, dis[v]);
    for (int u : adj[v]) {
      if (!vis[u]) {
        vis[u] = 1;
        q.push(u);
        if (cnt[u] == 1 && dis[v]) dis[u] = max(dis[u], dis[v] + 1);
      }
    }
  }
  cout << ans << '\n';
  for (int i = 0; i < n; ++i) {
    if (dis[i] == ans) {
      while (--ans) {
        cout << i + 1 << ' ';
        i = *adj2[i].begin();
      }
      cout << i + 1;
      break;
    }
  }
  return 0;
}

```