#include <bits/stdc++.h>
using namespace std;
queue<int> q;
bool type[1000000];
int ski[1000000];
int dist[1000000];
int from[1000000];
vector<int> bian[1000000];
int coun[1000000];
int n;
int ans = -1, pos;
void search() {
  for (int i = 1; i <= n; i++)
    if (type[i] == 1) q.push(i);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (coun[x] > 1) continue;
    if (ans < dist[x] && x) {
      ans = dist[x];
      pos = x;
    }
    for (int i = 0; i < bian[x].size(); i++)
      if (dist[bian[x][i]] < dist[x] + 1) {
        dist[bian[x][i]] = dist[x] + 1;
        from[bian[x][i]] = x;
        q.push(bian[x][i]);
      }
  }
}
int main(int argc, char *argv[]) {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> type[i];
  for (int i = 1; i <= n; i++) {
    int u;
    cin >> u;
    coun[u]++;
    bian[i].push_back(u);
  }
  search();
  cout << ans + 1 << endl;
  while (pos) {
    cout << pos << " ";
    pos = from[pos];
  }
  cout << endl;
  return EXIT_SUCCESS;
}

### 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;
queue<int> q;
bool type[1000000];
int ski[1000000];
int dist[1000000];
int from[1000000];
vector<int> bian[1000000];
int coun[1000000];
int n;
int ans = -1, pos;
void search() {
  for (int i = 1; i <= n; i++)
    if (type[i] == 1) q.push(i);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (coun[x] > 1) continue;
    if (ans < dist[x] && x) {
      ans = dist[x];
      pos = x;
    }
    for (int i = 0; i < bian[x].size(); i++)
      if (dist[bian[x][i]] < dist[x] + 1) {
        dist[bian[x][i]] = dist[x] + 1;
        from[bian[x][i]] = x;
        q.push(bian[x][i]);
      }
  }
}
int main(int argc, char *argv[]) {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> type[i];
  for (int i = 1; i <= n; i++) {
    int u;
    cin >> u;
    coun[u]++;
    bian[i].push_back(u);
  }
  search();
  cout << ans + 1 << endl;
  while (pos) {
    cout << pos << " ";
    pos = from[pos];
  }
  cout << endl;
  return EXIT_SUCCESS;
}

```