#include <bits/stdc++.h>
using namespace std;
const int mxn = 100005;
int t[mxn];
int p[mxn];
vector<int> g[mxn];
int main() {
  int n;
  cin >> n;
  int dd = 0;
  if (dd == 1) dd = 2;
  int maxl = 0;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
    maxl += t[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    g[p[i]].push_back(i);
  }
  int mlen = 1;
  int vis[mxn] = {0};
  int ans = -1;
  int ln[mxn] = {0};
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    if (!t[i] and g[i].size() > 1) continue;
    int s = i;
    int l = 1;
    vector<int> loc;
    loc.push_back(s);
    vis[i] = 1;
    while (p[s] != 0 and !vis[p[s]] and t[p[s]] != 1 and g[p[s]].size() <= 1) {
      s = p[s];
      loc.push_back(s);
      vis[s] = 1;
      l++;
    }
    if (p[s] != 0 and vis[p[s]] and !t[p[s]] and g[p[s]].size() <= 1)
      l += ln[p[s]];
    for (int ii = 0; ii < loc.size(); ii++) {
      ln[loc[ii]] = max(l - ii, ln[loc[ii]]);
    }
    if (t[i]) {
      if (mlen <= l) {
        mlen = l;
        ans = i;
      }
    }
  }
  cout << mlen << endl;
  vector<int> path;
  int s = ans;
  path.push_back(s);
  while (p[s] != 0 and g[p[s]].size() <= 1) {
    s = p[s];
    path.push_back(s);
  }
  reverse(path.begin(), path.end());
  for (int i = 0; i < path.size(); i++) {
    cout << path[i] << " ";
  }
}

### 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 mxn = 100005;
int t[mxn];
int p[mxn];
vector<int> g[mxn];
int main() {
  int n;
  cin >> n;
  int dd = 0;
  if (dd == 1) dd = 2;
  int maxl = 0;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
    maxl += t[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    g[p[i]].push_back(i);
  }
  int mlen = 1;
  int vis[mxn] = {0};
  int ans = -1;
  int ln[mxn] = {0};
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    if (!t[i] and g[i].size() > 1) continue;
    int s = i;
    int l = 1;
    vector<int> loc;
    loc.push_back(s);
    vis[i] = 1;
    while (p[s] != 0 and !vis[p[s]] and t[p[s]] != 1 and g[p[s]].size() <= 1) {
      s = p[s];
      loc.push_back(s);
      vis[s] = 1;
      l++;
    }
    if (p[s] != 0 and vis[p[s]] and !t[p[s]] and g[p[s]].size() <= 1)
      l += ln[p[s]];
    for (int ii = 0; ii < loc.size(); ii++) {
      ln[loc[ii]] = max(l - ii, ln[loc[ii]]);
    }
    if (t[i]) {
      if (mlen <= l) {
        mlen = l;
        ans = i;
      }
    }
  }
  cout << mlen << endl;
  vector<int> path;
  int s = ans;
  path.push_back(s);
  while (p[s] != 0 and g[p[s]].size() <= 1) {
    s = p[s];
    path.push_back(s);
  }
  reverse(path.begin(), path.end());
  for (int i = 0; i < path.size(); i++) {
    cout << path[i] << " ";
  }
}

```