#include <bits/stdc++.h>
using namespace std;
int pel(string s) {
  string t;
  t = s;
  reverse(t.begin(), t.end());
  if (s == t) return 1;
  return 0;
}
string toString(int n) {
  ostringstream ost;
  ost << n;
  ost.flush();
  return ost.str();
}
int toInt(string s) {
  int r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
bool isprime(int m) {
  if (m < 2) return 0;
  for (int i = 2; i * i <= m; i++)
    if (m % i == 0) return 0;
  return 1;
  return 0;
}
struct data {
  long long value;
  long long index;
};
int A[100005];
int B[100005];
int du[100005];
int ans[100005];
int result;
bool vis[100005];
int DFS(int x, int dep) {
  if (A[B[x]] || B[x] == 0 || du[B[x]] > 1) {
    if (dep > result) {
      result = dep;
      ans[dep] = x;
      return 1;
    }
    return 0;
  }
  if (DFS(B[x], dep + 1)) {
    ans[dep] = x;
    return 1;
  }
  return 0;
}
int main() {
  int n;
  int i;
  while (cin >> n) {
    memset(du, 0, sizeof(du));
    memset(vis, 0, sizeof(vis));
    for (i = 1; i < n + 1; i++) cin >> A[i];
    for (i = 1; i < n + 1; i++) {
      cin >> B[i];
      ++du[B[i]];
    }
    result = 0;
    for (i = 1; i < n + 1; i++) {
      if (A[i]) DFS(i, 1);
    }
    cout << result << "\n";
    for (int i = result; i >= 1; i--) {
      cout << ans[i] << " ";
    }
    cout << "\n";
  }
}

### Prompt

Develop a solution in cpp 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;
int pel(string s) {
  string t;
  t = s;
  reverse(t.begin(), t.end());
  if (s == t) return 1;
  return 0;
}
string toString(int n) {
  ostringstream ost;
  ost << n;
  ost.flush();
  return ost.str();
}
int toInt(string s) {
  int r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
bool isprime(int m) {
  if (m < 2) return 0;
  for (int i = 2; i * i <= m; i++)
    if (m % i == 0) return 0;
  return 1;
  return 0;
}
struct data {
  long long value;
  long long index;
};
int A[100005];
int B[100005];
int du[100005];
int ans[100005];
int result;
bool vis[100005];
int DFS(int x, int dep) {
  if (A[B[x]] || B[x] == 0 || du[B[x]] > 1) {
    if (dep > result) {
      result = dep;
      ans[dep] = x;
      return 1;
    }
    return 0;
  }
  if (DFS(B[x], dep + 1)) {
    ans[dep] = x;
    return 1;
  }
  return 0;
}
int main() {
  int n;
  int i;
  while (cin >> n) {
    memset(du, 0, sizeof(du));
    memset(vis, 0, sizeof(vis));
    for (i = 1; i < n + 1; i++) cin >> A[i];
    for (i = 1; i < n + 1; i++) {
      cin >> B[i];
      ++du[B[i]];
    }
    result = 0;
    for (i = 1; i < n + 1; i++) {
      if (A[i]) DFS(i, 1);
    }
    cout << result << "\n";
    for (int i = result; i >= 1; i--) {
      cout << ans[i] << " ";
    }
    cout << "\n";
  }
}

```