#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100 * 1000 + 23;
int n, t[MAXN], p[MAXN], od[MAXN];
queue<int> q;
pair<int, int> dp[MAXN], ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> t[i];
  for (int i = 0; i < n; i++) cin >> p[i], p[i]--, od[p[i]]++;
  for (int i = 0; i < n; i++)
    if (t[i]) q.push(i);
  while (!q.empty()) {
    if (~p[q.front()] && od[p[q.front()]] == 1) {
      dp[p[q.front()]] =
          max(dp[p[q.front()]], {dp[q.front()].first + 1, q.front() + 1});
      q.push(p[q.front()]);
    }
    q.pop();
  }
  int f = -1;
  for (int i = 0; i < n; i++) {
    if ((t[i] == 1 || dp[i].second) && ans.first <= dp[i].first) {
      ans = dp[i];
      f = i + 1;
    }
  }
  cout << ans.first + 1 << '\n' << f << ' ';
  while (ans.second) {
    cout << ans.second << ' ';
    ans = dp[ans.second - 1];
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int MAXN = 100 * 1000 + 23;
int n, t[MAXN], p[MAXN], od[MAXN];
queue<int> q;
pair<int, int> dp[MAXN], ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> t[i];
  for (int i = 0; i < n; i++) cin >> p[i], p[i]--, od[p[i]]++;
  for (int i = 0; i < n; i++)
    if (t[i]) q.push(i);
  while (!q.empty()) {
    if (~p[q.front()] && od[p[q.front()]] == 1) {
      dp[p[q.front()]] =
          max(dp[p[q.front()]], {dp[q.front()].first + 1, q.front() + 1});
      q.push(p[q.front()]);
    }
    q.pop();
  }
  int f = -1;
  for (int i = 0; i < n; i++) {
    if ((t[i] == 1 || dp[i].second) && ans.first <= dp[i].first) {
      ans = dp[i];
      f = i + 1;
    }
  }
  cout << ans.first + 1 << '\n' << f << ' ';
  while (ans.second) {
    cout << ans.second << ' ';
    ans = dp[ans.second - 1];
  }
  return 0;
}

```