#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, inf = 1e9;
int n;
vector<int> adj[N];
int dp[N], parent[N];
;
int solve(int v, int par = -1) {
  if (dp[v] != -1) return dp[v];
  dp[v] = -inf;
  int deg = adj[v].size() - (par != -1);
  if (deg != 1) return dp[v];
  for (auto u : adj[v]) {
    int x = solve(u) + 1;
    if (x > dp[v]) parent[v] = u, dp[v] = x;
  }
  return dp[v];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  memset(dp, -1, sizeof dp);
  memset(parent, -1, sizeof parent);
  for (int i = 0; i < n; i++) {
    int type;
    cin >> type;
    if (type == 1) dp[i] = 1;
  }
  for (int i = 0; i < n; i++) {
    int par;
    cin >> par;
    par--;
    if (par >= 0) adj[par].push_back(i);
  }
  int ans = 0, st = -1;
  for (int i = 0; i < n; i++) {
    if (ans < solve(i)) {
      st = i;
      ans = solve(i);
    }
  }
  cout << ans << "\n";
  while (st != -1) {
    cout << st + 1 << " ";
    st = parent[st];
  }
  cout << "\n";
}

### 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 = 1e5 + 5, inf = 1e9;
int n;
vector<int> adj[N];
int dp[N], parent[N];
;
int solve(int v, int par = -1) {
  if (dp[v] != -1) return dp[v];
  dp[v] = -inf;
  int deg = adj[v].size() - (par != -1);
  if (deg != 1) return dp[v];
  for (auto u : adj[v]) {
    int x = solve(u) + 1;
    if (x > dp[v]) parent[v] = u, dp[v] = x;
  }
  return dp[v];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  memset(dp, -1, sizeof dp);
  memset(parent, -1, sizeof parent);
  for (int i = 0; i < n; i++) {
    int type;
    cin >> type;
    if (type == 1) dp[i] = 1;
  }
  for (int i = 0; i < n; i++) {
    int par;
    cin >> par;
    par--;
    if (par >= 0) adj[par].push_back(i);
  }
  int ans = 0, st = -1;
  for (int i = 0; i < n; i++) {
    if (ans < solve(i)) {
      st = i;
      ans = solve(i);
    }
  }
  cout << ans << "\n";
  while (st != -1) {
    cout << st + 1 << " ";
    st = parent[st];
  }
  cout << "\n";
}

```