#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
T min(T a, U b) {
  if (a < (U)b) return a;
  return b;
}
template <class T, class U>
T max(T a, U b) {
  if (a < (U)b) return b;
  return a;
}
const int INF = 1000000000;
const long long MOD = 1000000009;
const double eps = 1e-9;
vector<vector<int> > g;
vector<char> used;
vector<int> p;
vector<int> h;
vector<int> sz;
void dfs(int v, int curh = 0, int par = -1) {
  used[v] = true;
  h[v] = curh;
  p[v] = par;
  for (int i = 0; i < g[v].size(); ++i) {
    if (!used[g[v][i]] && sz[g[v][i]] <= 1) dfs(g[v][i], curh + 1, v);
  }
}
int main() {
  int N;
  cin >> N;
  g.resize(N);
  used.assign(N, false);
  h.assign(N, 0);
  p.assign(N, -1);
  sz.assign(N, 0);
  int type[N];
  for (int i = 0; i < N; ++i) {
    cin >> type[i];
  }
  int curp[N];
  for (int i = 0; i < N; ++i) {
    cin >> curp[i];
    curp[i]--;
    if (curp[i] >= 0) {
      sz[curp[i]]++;
      g[i].push_back(curp[i]);
    }
  }
  for (int i = 0; i < N; ++i) {
    if (type[i]) {
      dfs(i);
    }
  }
  int maxh = -1, maxid = -1;
  for (int i = 0; i < N; ++i) {
    if (h[i] > maxh && used[i]) {
      maxh = h[i];
      maxid = i;
    }
  }
  vector<int> res;
  for (int i = maxid; i != -1; i = p[i]) {
    res.push_back(i + 1);
  }
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); ++i) cout << res[i] << " ";
  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;
template <class T, class U>
T min(T a, U b) {
  if (a < (U)b) return a;
  return b;
}
template <class T, class U>
T max(T a, U b) {
  if (a < (U)b) return b;
  return a;
}
const int INF = 1000000000;
const long long MOD = 1000000009;
const double eps = 1e-9;
vector<vector<int> > g;
vector<char> used;
vector<int> p;
vector<int> h;
vector<int> sz;
void dfs(int v, int curh = 0, int par = -1) {
  used[v] = true;
  h[v] = curh;
  p[v] = par;
  for (int i = 0; i < g[v].size(); ++i) {
    if (!used[g[v][i]] && sz[g[v][i]] <= 1) dfs(g[v][i], curh + 1, v);
  }
}
int main() {
  int N;
  cin >> N;
  g.resize(N);
  used.assign(N, false);
  h.assign(N, 0);
  p.assign(N, -1);
  sz.assign(N, 0);
  int type[N];
  for (int i = 0; i < N; ++i) {
    cin >> type[i];
  }
  int curp[N];
  for (int i = 0; i < N; ++i) {
    cin >> curp[i];
    curp[i]--;
    if (curp[i] >= 0) {
      sz[curp[i]]++;
      g[i].push_back(curp[i]);
    }
  }
  for (int i = 0; i < N; ++i) {
    if (type[i]) {
      dfs(i);
    }
  }
  int maxh = -1, maxid = -1;
  for (int i = 0; i < N; ++i) {
    if (h[i] > maxh && used[i]) {
      maxh = h[i];
      maxid = i;
    }
  }
  vector<int> res;
  for (int i = maxid; i != -1; i = p[i]) {
    res.push_back(i + 1);
  }
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); ++i) cout << res[i] << " ";
  return 0;
}

```