#include <bits/stdc++.h>
using namespace std;
const long double pi = acos(-1.0);
const long long mod7 = 1000000007;
const long long mod9 = 1000000009;
const int linf = 2000000000;
const long long llinf = 1000000000000000000ll;
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
pair<int, int> a[1000001];
vector<int> ans;
vector<int> gr[100001];
int p[100001];
int dep[100001];
bool used[100001];
int dfs(int v) {
  int cur = -linf;
  if (!used[v]) {
    used[v] = true;
    if (gr[v].size() <= 1)
      for (int i = 0; i < gr[v].size(); ++i) {
        int to = gr[v][i];
        if (a[to].first == 1) {
          cur = 1;
          p[v] = to;
        } else if (dep[to] >= 0) {
          if (dep[to] == linf) continue;
          cur = dep[to] + 1;
          p[v] = to;
        } else {
          dfs(to);
          if (dep[to] == linf) continue;
          cur = dep[to] + 1;
          p[v] = to;
        }
      }
  }
  if (cur == -linf)
    dep[v] = linf;
  else
    dep[v] = cur;
  used[v] = false;
  return cur;
}
int main() {
  int n, m, x;
  cin >> n;
  int best = 0;
  int d = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].first;
    if (!a[i].first)
      dep[i] = -linf;
    else
      best = i;
    p[i] = -1;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].second;
    if (a[i].second) {
      gr[a[i].second].push_back(i);
    }
  }
  for (int i = 1; i <= n; ++i) {
    int k = dfs(i);
    if (k > d) {
      d = k;
      best = i;
    }
  }
  cout << d + 1 << endl;
  while (p[best] != -1) {
    cout << best << ' ';
    best = p[best];
  }
  cout << best;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const long double pi = acos(-1.0);
const long long mod7 = 1000000007;
const long long mod9 = 1000000009;
const int linf = 2000000000;
const long long llinf = 1000000000000000000ll;
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
pair<int, int> a[1000001];
vector<int> ans;
vector<int> gr[100001];
int p[100001];
int dep[100001];
bool used[100001];
int dfs(int v) {
  int cur = -linf;
  if (!used[v]) {
    used[v] = true;
    if (gr[v].size() <= 1)
      for (int i = 0; i < gr[v].size(); ++i) {
        int to = gr[v][i];
        if (a[to].first == 1) {
          cur = 1;
          p[v] = to;
        } else if (dep[to] >= 0) {
          if (dep[to] == linf) continue;
          cur = dep[to] + 1;
          p[v] = to;
        } else {
          dfs(to);
          if (dep[to] == linf) continue;
          cur = dep[to] + 1;
          p[v] = to;
        }
      }
  }
  if (cur == -linf)
    dep[v] = linf;
  else
    dep[v] = cur;
  used[v] = false;
  return cur;
}
int main() {
  int n, m, x;
  cin >> n;
  int best = 0;
  int d = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].first;
    if (!a[i].first)
      dep[i] = -linf;
    else
      best = i;
    p[i] = -1;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].second;
    if (a[i].second) {
      gr[a[i].second].push_back(i);
    }
  }
  for (int i = 1; i <= n; ++i) {
    int k = dfs(i);
    if (k > d) {
      d = k;
      best = i;
    }
  }
  cout << d + 1 << endl;
  while (p[best] != -1) {
    cout << best << ' ';
    best = p[best];
  }
  cout << best;
  return 0;
}

```