#include <bits/stdc++.h>
using namespace std;
typedef struct tag {
  int status;
  int parent;
  int degree = 0;
} node;
node travel[100005];
int use[100005];
bool compare(node a, node b) { return (a.status > b.status); }
vector<int> tmp, path;
vector<int>::iterator it;
void dfs(int ind) {
  if (travel[ind].degree >= 2 || ind == 0) return;
  tmp.push_back(ind);
  dfs(travel[ind].parent);
}
int main() {
  int n, max = INT_MIN, temp, k = 0;
  scanf("%d", &n);
  for (int j = 1; j <= n; j++) {
    scanf("%d", &travel[j].status);
    if (travel[j].status) {
      use[k++] = j;
    }
  }
  for (int j = 1; j <= n; j++) {
    scanf("%d", &travel[j].parent);
    travel[travel[j].parent].degree++;
  }
  for (int i = 0; i < k; i++) {
    dfs(use[i]);
    if (tmp.size() > path.size()) {
      path.clear();
      it = path.begin();
      path.insert(it, tmp.rbegin(), tmp.rend());
    }
    tmp.clear();
  }
  printf("%d\n", path.size());
  for (it = path.begin(); it != path.end(); it++) printf("%d ", *it);
  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;
typedef struct tag {
  int status;
  int parent;
  int degree = 0;
} node;
node travel[100005];
int use[100005];
bool compare(node a, node b) { return (a.status > b.status); }
vector<int> tmp, path;
vector<int>::iterator it;
void dfs(int ind) {
  if (travel[ind].degree >= 2 || ind == 0) return;
  tmp.push_back(ind);
  dfs(travel[ind].parent);
}
int main() {
  int n, max = INT_MIN, temp, k = 0;
  scanf("%d", &n);
  for (int j = 1; j <= n; j++) {
    scanf("%d", &travel[j].status);
    if (travel[j].status) {
      use[k++] = j;
    }
  }
  for (int j = 1; j <= n; j++) {
    scanf("%d", &travel[j].parent);
    travel[travel[j].parent].degree++;
  }
  for (int i = 0; i < k; i++) {
    dfs(use[i]);
    if (tmp.size() > path.size()) {
      path.clear();
      it = path.begin();
      path.insert(it, tmp.rbegin(), tmp.rend());
    }
    tmp.clear();
  }
  printf("%d\n", path.size());
  for (it = path.begin(); it != path.end(); it++) printf("%d ", *it);
  return (0);
}

```