#include <bits/stdc++.h>
using namespace std;
int n, type[100001], a[100001], lead[100001], cur[100001];
bool vs[100001];
void input() {
  scanf("%d", &n);
  for (int i = 1, _b = n; i <= _b; i++) scanf("%d", &type[i]);
  int x;
  memset(lead, 0, sizeof(lead));
  memset(a, 0, sizeof(a));
  for (int i = 1, _b = n; i <= _b; i++) {
    scanf("%d", &x);
    a[x] = i;
    lead[x]++;
  }
}
void dfs(int i) {
  vs[i] = true;
  if (type[i]) {
    cur[i] = 1;
    return;
  }
  if (lead[a[i]] != 1 && !type[a[i]]) return;
  if (!vs[a[i]]) dfs(a[i]);
  if (cur[a[i]]) cur[i] = cur[a[i]] + 1;
}
void process() {
  memset(vs, false, sizeof(vs));
  memset(cur, 0, sizeof(cur));
  for (int i = 0, _b = n; i <= _b; i++)
    if (lead[i] != 1 && !type[i]) vs[i] = true;
  for (int i = 1, _b = n; i <= _b; i++) {
    if (!vs[i]) dfs(i);
  }
  int ans = 0;
  int st = 0;
  for (int i = 1, _b = n; i <= _b; i++) {
    if (ans < cur[i]) {
      ans = cur[i];
      st = i;
    }
  }
  cout << ans << endl;
  while (!type[st]) {
    printf("%d ", st);
    st = a[st];
  }
  cout << st;
}
int main() {
  input();
  process();
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int n, type[100001], a[100001], lead[100001], cur[100001];
bool vs[100001];
void input() {
  scanf("%d", &n);
  for (int i = 1, _b = n; i <= _b; i++) scanf("%d", &type[i]);
  int x;
  memset(lead, 0, sizeof(lead));
  memset(a, 0, sizeof(a));
  for (int i = 1, _b = n; i <= _b; i++) {
    scanf("%d", &x);
    a[x] = i;
    lead[x]++;
  }
}
void dfs(int i) {
  vs[i] = true;
  if (type[i]) {
    cur[i] = 1;
    return;
  }
  if (lead[a[i]] != 1 && !type[a[i]]) return;
  if (!vs[a[i]]) dfs(a[i]);
  if (cur[a[i]]) cur[i] = cur[a[i]] + 1;
}
void process() {
  memset(vs, false, sizeof(vs));
  memset(cur, 0, sizeof(cur));
  for (int i = 0, _b = n; i <= _b; i++)
    if (lead[i] != 1 && !type[i]) vs[i] = true;
  for (int i = 1, _b = n; i <= _b; i++) {
    if (!vs[i]) dfs(i);
  }
  int ans = 0;
  int st = 0;
  for (int i = 1, _b = n; i <= _b; i++) {
    if (ans < cur[i]) {
      ans = cur[i];
      st = i;
    }
  }
  cout << ans << endl;
  while (!type[st]) {
    printf("%d ", st);
    st = a[st];
  }
  cout << st;
}
int main() {
  input();
  process();
  return 0;
}

```