#include <bits/stdc++.h>
using namespace std;
int toInt(string s) {
  int r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
long long toInt64(string s) {
  long long r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
double toDouble(string s) {
  double r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
string toString(long long n) {
  string s, s1;
  while (n / 10 > 0) {
    s += (char)((n % 10) + 48);
    n /= 10;
  }
  s += (char)((n % 10) + 48);
  n /= 10;
  s1 = s;
  for (long long i = 0; i < s.length(); i++) s1[(s.length() - 1) - i] = s[i];
  return s1;
}
bool isUpperCase(char c) { return c >= 'A' && c <= 'Z'; }
bool isLowerCase(char c) { return c >= 'a' && c <= 'z'; }
bool isLetter(char c) { return c >= 'A' && c <= 'Z' || c >= 'a' && c <= 'z'; }
bool isDigit(char c) { return c >= '0' && c <= '9'; }
char toLowerCase(char c) { return (isUpperCase(c)) ? (c + 32) : c; }
char toUpperCase(char c) { return (isLowerCase(c)) ? (c - 32) : c; }
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
bool grid[100001];
int masir[100001];
int yek[100001], Yek = 0;
short int how[100001];
vector<int> BEST;
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (a == 1)
      grid[i] = true;
    else
      grid[i] = false;
    if (grid[i]) yek[Yek++] = i;
  }
  for (long long i = 0; i < n; i++) {
    cin >> masir[i];
    masir[i]--;
    if (how[masir[i]] != 2) how[masir[i]]++;
  }
  int Max = 0;
  for (long long i = 0; i < Yek; i++) {
    int Masir[100001], count = 0, dest = yek[i];
    Masir[count++] = dest;
    while (masir[dest] != -1) {
      dest = masir[dest];
      if (grid[dest] || how[dest] >= 2) break;
      Masir[count++] = dest;
    }
    if (Max < count) {
      BEST.clear();
      for (int j = count - 1; j >= 0; j--) BEST.push_back(Masir[j]);
      Max = count;
    }
  }
  int SIZE = BEST.size();
  cout << SIZE << endl;
  for (long long i = 0; i < SIZE - 1; i++) cout << BEST[i] + 1 << ' ';
  cout << BEST[SIZE - 1] + 1 << endl;
}

### Prompt

Your challenge is to write 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;
int toInt(string s) {
  int r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
long long toInt64(string s) {
  long long r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
double toDouble(string s) {
  double r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
string toString(long long n) {
  string s, s1;
  while (n / 10 > 0) {
    s += (char)((n % 10) + 48);
    n /= 10;
  }
  s += (char)((n % 10) + 48);
  n /= 10;
  s1 = s;
  for (long long i = 0; i < s.length(); i++) s1[(s.length() - 1) - i] = s[i];
  return s1;
}
bool isUpperCase(char c) { return c >= 'A' && c <= 'Z'; }
bool isLowerCase(char c) { return c >= 'a' && c <= 'z'; }
bool isLetter(char c) { return c >= 'A' && c <= 'Z' || c >= 'a' && c <= 'z'; }
bool isDigit(char c) { return c >= '0' && c <= '9'; }
char toLowerCase(char c) { return (isUpperCase(c)) ? (c + 32) : c; }
char toUpperCase(char c) { return (isLowerCase(c)) ? (c - 32) : c; }
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
bool grid[100001];
int masir[100001];
int yek[100001], Yek = 0;
short int how[100001];
vector<int> BEST;
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (a == 1)
      grid[i] = true;
    else
      grid[i] = false;
    if (grid[i]) yek[Yek++] = i;
  }
  for (long long i = 0; i < n; i++) {
    cin >> masir[i];
    masir[i]--;
    if (how[masir[i]] != 2) how[masir[i]]++;
  }
  int Max = 0;
  for (long long i = 0; i < Yek; i++) {
    int Masir[100001], count = 0, dest = yek[i];
    Masir[count++] = dest;
    while (masir[dest] != -1) {
      dest = masir[dest];
      if (grid[dest] || how[dest] >= 2) break;
      Masir[count++] = dest;
    }
    if (Max < count) {
      BEST.clear();
      for (int j = count - 1; j >= 0; j--) BEST.push_back(Masir[j]);
      Max = count;
    }
  }
  int SIZE = BEST.size();
  cout << SIZE << endl;
  for (long long i = 0; i < SIZE - 1; i++) cout << BEST[i] + 1 << ' ';
  cout << BEST[SIZE - 1] + 1 << endl;
}

```