#include <bits/stdc++.h>
using namespace std;
inline void debug() {}
template <typename _type, typename... type>
inline void debug(_type arg, const type&... next) {}
vector<int> adj[100006];
int prevv[100006], n;
queue<int> que;
bool hotel[100006];
int ile[100006], tab[100006];
int main() {
  scanf("%d", &n);
  int a;
  debug("bka");
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    if (a) {
      hotel[i] = true;
      que.push(i);
      debug(i);
    }
    prevv[i] = -1;
  }
  for (int(i) = 0; (i) < (n); (i)++) {
    scanf("%d", &tab[i]);
    ile[tab[i]]++;
  }
  for (int(i) = 0; (i) < (n); (i)++) {
    if (tab[i] && ile[tab[i]] < 2) {
      adj[i].push_back(tab[i] - 1);
    }
  }
  debug("and");
  while (!que.empty()) {
    a = que.front();
    debug(a);
    que.pop();
    for (int(i) = 0; (i) < (adj[a].size()); (i)++) {
      que.push(adj[a][i]);
      prevv[adj[a][i]] = a;
      debug("prev[", adj[a][i], " = ", a);
      adj[a].clear();
    }
  }
  debug("last = ", a);
  vector<int> res;
  res.push_back(a);
  debug("herE");
  while (!hotel[a]) {
    a = prevv[a];
    res.push_back(a);
  }
  printf("%d\n", res.size());
  for (int(i) = 0; (i) < (res.size()); (i)++) printf("%d ", res[i] + 1);
  printf("\n");
}

### Prompt

Generate a cpp solution to 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;
inline void debug() {}
template <typename _type, typename... type>
inline void debug(_type arg, const type&... next) {}
vector<int> adj[100006];
int prevv[100006], n;
queue<int> que;
bool hotel[100006];
int ile[100006], tab[100006];
int main() {
  scanf("%d", &n);
  int a;
  debug("bka");
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    if (a) {
      hotel[i] = true;
      que.push(i);
      debug(i);
    }
    prevv[i] = -1;
  }
  for (int(i) = 0; (i) < (n); (i)++) {
    scanf("%d", &tab[i]);
    ile[tab[i]]++;
  }
  for (int(i) = 0; (i) < (n); (i)++) {
    if (tab[i] && ile[tab[i]] < 2) {
      adj[i].push_back(tab[i] - 1);
    }
  }
  debug("and");
  while (!que.empty()) {
    a = que.front();
    debug(a);
    que.pop();
    for (int(i) = 0; (i) < (adj[a].size()); (i)++) {
      que.push(adj[a][i]);
      prevv[adj[a][i]] = a;
      debug("prev[", adj[a][i], " = ", a);
      adj[a].clear();
    }
  }
  debug("last = ", a);
  vector<int> res;
  res.push_back(a);
  debug("herE");
  while (!hotel[a]) {
    a = prevv[a];
    res.push_back(a);
  }
  printf("%d\n", res.size());
  for (int(i) = 0; (i) < (res.size()); (i)++) printf("%d ", res[i] + 1);
  printf("\n");
}

```