#include <bits/stdc++.h>
using namespace std;
int n;
vector<vector<int>> Pre;
vector<int> Next;
vector<int> target;
vector<int> res;
void dfs(int v, vector<int> &tmp) {
  tmp.push_back(v);
  if (Pre[v].empty()) {
    if (tmp.size() > res.size()) res = tmp;
    tmp.pop_back();
    return;
  }
  for (auto u : Pre[v]) {
    if (Next[u] != 1) {
      if (tmp.size() > res.size()) res = vector<int>(tmp);
    } else {
      dfs(u, tmp);
    }
  }
  tmp.pop_back();
}
int main() {
  cin >> n;
  Pre = vector<vector<int>>(n + 1);
  Next = vector<int>(n + 1, 0);
  target = vector<int>(n + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &target[i]);
  }
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d", &p);
    if (p != 0) {
      Next[p]++;
      Pre[i].push_back(p);
    }
  }
  vector<int> tmp;
  for (int i = 1; i <= n; i++) {
    if (target[i]) {
      dfs(i, tmp);
    }
  }
  printf("%d\n", res.size());
  for (int i = res.size() - 1; i >= 0; i--) {
    printf("%d", res[i]);
    if (i != 0) printf(" ");
  }
  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>
using namespace std;
int n;
vector<vector<int>> Pre;
vector<int> Next;
vector<int> target;
vector<int> res;
void dfs(int v, vector<int> &tmp) {
  tmp.push_back(v);
  if (Pre[v].empty()) {
    if (tmp.size() > res.size()) res = tmp;
    tmp.pop_back();
    return;
  }
  for (auto u : Pre[v]) {
    if (Next[u] != 1) {
      if (tmp.size() > res.size()) res = vector<int>(tmp);
    } else {
      dfs(u, tmp);
    }
  }
  tmp.pop_back();
}
int main() {
  cin >> n;
  Pre = vector<vector<int>>(n + 1);
  Next = vector<int>(n + 1, 0);
  target = vector<int>(n + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &target[i]);
  }
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d", &p);
    if (p != 0) {
      Next[p]++;
      Pre[i].push_back(p);
    }
  }
  vector<int> tmp;
  for (int i = 1; i <= n; i++) {
    if (target[i]) {
      dfs(i, tmp);
    }
  }
  printf("%d\n", res.size());
  for (int i = res.size() - 1; i >= 0; i--) {
    printf("%d", res[i]);
    if (i != 0) printf(" ");
  }
  return 0;
}

```