#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 123;
const int mod = 1e9 + 7;
const int INF = 1e9 + 1;
const long long INFL = 3e18 + 1;
const double pi = acos(-1.0);
int n, is[N], p[N];
vector<int> comp, ans;
int mx;
int cnt[N];
void dfs(int v) {
  comp.push_back(v);
  if (cnt[p[v]] == 1 && !is[p[v]]) dfs(p[v]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> is[i];
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] != 0)
      cnt[p[i]]++;
    else
      cnt[p[i]] = INF;
  }
  for (int i = 1; i <= n; i++) {
    if (is[i] == 1) {
      comp.clear();
      dfs(i);
      if (comp.size() > mx) {
        mx = comp.size();
        ans = comp;
      }
    }
  }
  cout << ans.size() << endl;
  reverse(ans.begin(), ans.end());
  for (auto i : ans) cout << 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;
const int N = 2e5 + 123;
const int mod = 1e9 + 7;
const int INF = 1e9 + 1;
const long long INFL = 3e18 + 1;
const double pi = acos(-1.0);
int n, is[N], p[N];
vector<int> comp, ans;
int mx;
int cnt[N];
void dfs(int v) {
  comp.push_back(v);
  if (cnt[p[v]] == 1 && !is[p[v]]) dfs(p[v]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> is[i];
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] != 0)
      cnt[p[i]]++;
    else
      cnt[p[i]] = INF;
  }
  for (int i = 1; i <= n; i++) {
    if (is[i] == 1) {
      comp.clear();
      dfs(i);
      if (comp.size() > mx) {
        mx = comp.size();
        ans = comp;
      }
    }
  }
  cout << ans.size() << endl;
  reverse(ans.begin(), ans.end());
  for (auto i : ans) cout << i << ' ';
  return 0;
}

```