#include <bits/stdc++.h>
using namespace std;
int morh[100001], parent[100001], pos[100001], temp = 0, ans = 0;
int calculate(int i) {
  int posi = i, ans = 0;
  while (posi != 0) {
    if (pos[posi] > 1)
      break;
    else {
      posi = parent[posi];
      ans++;
    }
  }
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &morh[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &parent[i]);
    pos[parent[i]]++;
  }
  int temp2;
  for (int i = 1; i <= n; i++) {
    if (morh[i] == 1) {
      temp2 = calculate(i);
      if (temp2 > ans) {
        ans = temp2;
        temp = i;
      }
    }
  }
  temp2 = ans;
  int give_ans[ans], i = 0;
  for (int j = 0; j < temp2; j++) {
    give_ans[i] = temp;
    i++;
    temp = parent[temp];
  }
  printf("%d\n", ans);
  for (int i = ans - 1; i >= 0; i--) printf("%d ", give_ans[i]);
  printf("\n");
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 morh[100001], parent[100001], pos[100001], temp = 0, ans = 0;
int calculate(int i) {
  int posi = i, ans = 0;
  while (posi != 0) {
    if (pos[posi] > 1)
      break;
    else {
      posi = parent[posi];
      ans++;
    }
  }
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &morh[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &parent[i]);
    pos[parent[i]]++;
  }
  int temp2;
  for (int i = 1; i <= n; i++) {
    if (morh[i] == 1) {
      temp2 = calculate(i);
      if (temp2 > ans) {
        ans = temp2;
        temp = i;
      }
    }
  }
  temp2 = ans;
  int give_ans[ans], i = 0;
  for (int j = 0; j < temp2; j++) {
    give_ans[i] = temp;
    i++;
    temp = parent[temp];
  }
  printf("%d\n", ans);
  for (int i = ans - 1; i >= 0; i--) printf("%d ", give_ans[i]);
  printf("\n");
}

```