#include <bits/stdc++.h>
using namespace std;
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
int hotel[100001];
int track[100001];
int num[100001];
int sol[100001];
int main() {
  int i, j, k, n, m, res = 0, cnt = 0, mx = 0, mn = (1 << 30), x;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &hotel[i]);
    num[i] = 0;
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &track[i]);
    num[track[i] - 1]++;
    if (track[i] == i + 1) track[i] = 0;
  }
  for (i = 0; i < n; i++) {
    if (hotel[i]) {
      j = i;
      k = 0;
      while (track[j] && (hotel[j] == 0 || k == 0) && num[j] <= 1) {
        j = track[j] - 1;
        k++;
      }
      if (track[j] == 0 && (hotel[j] == 0 || k == 0) && num[j] <= 1) k++;
      if (k > mx) {
        mx = k;
        x = i;
      }
    }
  }
  printf("%d\n", mx);
  j = x;
  k = 0;
  while (track[j] && (hotel[j] == 0 || k == 0) && num[j] <= 1) {
    sol[k++] = j + 1;
    j = track[j] - 1;
  }
  if (track[j] == 0 && (hotel[j] == 0 || k == 0) && num[j] <= 1)
    sol[k++] = j + 1;
  for (i = k - 1; i >= 0; i--) {
    if (i == 0)
      printf("%d\n", sol[i]);
    else
      printf("%d ", sol[i]);
  }
  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;
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
int hotel[100001];
int track[100001];
int num[100001];
int sol[100001];
int main() {
  int i, j, k, n, m, res = 0, cnt = 0, mx = 0, mn = (1 << 30), x;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &hotel[i]);
    num[i] = 0;
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &track[i]);
    num[track[i] - 1]++;
    if (track[i] == i + 1) track[i] = 0;
  }
  for (i = 0; i < n; i++) {
    if (hotel[i]) {
      j = i;
      k = 0;
      while (track[j] && (hotel[j] == 0 || k == 0) && num[j] <= 1) {
        j = track[j] - 1;
        k++;
      }
      if (track[j] == 0 && (hotel[j] == 0 || k == 0) && num[j] <= 1) k++;
      if (k > mx) {
        mx = k;
        x = i;
      }
    }
  }
  printf("%d\n", mx);
  j = x;
  k = 0;
  while (track[j] && (hotel[j] == 0 || k == 0) && num[j] <= 1) {
    sol[k++] = j + 1;
    j = track[j] - 1;
  }
  if (track[j] == 0 && (hotel[j] == 0 || k == 0) && num[j] <= 1)
    sol[k++] = j + 1;
  for (i = k - 1; i >= 0; i--) {
    if (i == 0)
      printf("%d\n", sol[i]);
    else
      printf("%d ", sol[i]);
  }
  return 0;
}

```