#include <bits/stdc++.h>
int a[100003], to[100003], from[100003], c[100003], cc[100003], die[100003], n,
    i, j, k, ans;
int main() {
  while (~scanf("%d", &n)) {
    for (i = 1; i <= n; i++)
      scanf("%d", &a[i]), to[i] = 0, from[i] = 0, die[i] = 0;
    for (from[0] = 0, i = 1; i <= n; i++) {
      scanf("%d", &from[i]);
      if (to[from[i]] == 0)
        to[from[i]] = i;
      else
        die[from[i]] = 1, from[i] = 0;
    }
    for (ans = 0, i = 1; i <= n; i++) {
      if (a[i] == 1 && !die[i]) {
        cc[0] = i, j = from[i], k = 1;
        while (1) {
          if (j != 0 && a[j] == 0 && !die[j])
            cc[k++] = j, j = from[j];
          else
            break;
        }
        if (k > ans) {
          for (j = k - 1; j >= 0; j--) c[k - 1 - j] = cc[j];
          ans = k;
        }
      }
    }
    printf("%d\n", ans);
    for (i = 0; i < ans - 1; i++) printf("%d ", c[i]);
    printf("%d\n", c[ans - 1]);
  }
  return 0;
}

### 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>
int a[100003], to[100003], from[100003], c[100003], cc[100003], die[100003], n,
    i, j, k, ans;
int main() {
  while (~scanf("%d", &n)) {
    for (i = 1; i <= n; i++)
      scanf("%d", &a[i]), to[i] = 0, from[i] = 0, die[i] = 0;
    for (from[0] = 0, i = 1; i <= n; i++) {
      scanf("%d", &from[i]);
      if (to[from[i]] == 0)
        to[from[i]] = i;
      else
        die[from[i]] = 1, from[i] = 0;
    }
    for (ans = 0, i = 1; i <= n; i++) {
      if (a[i] == 1 && !die[i]) {
        cc[0] = i, j = from[i], k = 1;
        while (1) {
          if (j != 0 && a[j] == 0 && !die[j])
            cc[k++] = j, j = from[j];
          else
            break;
        }
        if (k > ans) {
          for (j = k - 1; j >= 0; j--) c[k - 1 - j] = cc[j];
          ans = k;
        }
      }
    }
    printf("%d\n", ans);
    for (i = 0; i < ans - 1; i++) printf("%d ", c[i]);
    printf("%d\n", c[ans - 1]);
  }
  return 0;
}

```