#include <bits/stdc++.h>
using namespace std;
const int sz = 1500000;
const long long MAX = 10000000000000000;
template <typename T>
T abs(T a) {
  return a < 0 ? -a : a;
}
const long long MOD = 1e9 + 7;
const long long MASK = 1099511627776LL;
int napr[8][2] = {{0, 1}, {1, 0},  {-1, 0}, {0, -1},
                  {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
const int PRIME = 4999999;
vector<long long> v;
vector<long double> b(401, -1);
vector<char> used(401, false);
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int len;
bool comparer(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first / len == b.first.first / len) {
    return a.first.second < b.first.second;
  }
  return a.first.first < b.first.first;
}
long long poww(long long a, long long step) {
  if (step == 0) {
    return 1;
  }
  if (step % 2) {
    return (a * poww(a, step - 1) % (int)(1e9 + 7));
  } else {
    long long buf = poww(a, step / 2);
    return (buf * buf) % (int)(1e9 + 7);
  }
}
vector<int> fenvik;
int n;
int maxV = -1;
vector<int> isOt;
vector<vector<int> > g, rg;
vector<int> flag;
vector<int> ts;
vector<int> path;
bool isOtv = true;
void dfs(int v, int len = 0) {
  flag[v] = 1;
  if (len > maxV) {
    maxV = max(len, maxV);
    path.clear();
    isOtv = false;
  }
  for (long long i = 0; i < g[v].size(); i++) {
    if (!flag[g[v][i]]) dfs(g[v][i], len + 1);
  }
  if (!isOtv) {
    path.push_back(v);
  }
}
int main() {
  int n;
  cin >> n;
  isOt.resize(n);
  g.resize(n);
  vector<int> start;
  flag.resize(n, 0);
  vector<int> otel;
  for (long long i = 0; i < n; i++) {
    cin >> isOt[i];
    if (isOt[i]) {
      otel.push_back(i);
    }
  }
  vector<int> cp(n, 0);
  for (long long i = 0; i < n; i++) {
    int buf;
    cin >> buf;
    if (buf) {
      buf--;
      cp[buf]++;
      if (cp[buf] > 1) flag[buf] = 1;
      g[i].push_back(buf);
    } else {
      start.push_back(i);
    }
  }
  flag.resize(n, 0);
  vector<int> res;
  for (long long i = 0; i < otel.size(); i++) {
    isOtv = true;
    path.clear();
    int oldV = maxV;
    dfs(otel[i]);
    if (oldV < maxV) {
      res = path;
      oldV = maxV;
    }
  }
  cout << res.size() << endl;
  for (long long i = 0; i < res.size(); i++) {
    cout << res[i] + 1 << ' ';
  }
  return 0;
}

### 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 sz = 1500000;
const long long MAX = 10000000000000000;
template <typename T>
T abs(T a) {
  return a < 0 ? -a : a;
}
const long long MOD = 1e9 + 7;
const long long MASK = 1099511627776LL;
int napr[8][2] = {{0, 1}, {1, 0},  {-1, 0}, {0, -1},
                  {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
const int PRIME = 4999999;
vector<long long> v;
vector<long double> b(401, -1);
vector<char> used(401, false);
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int len;
bool comparer(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first / len == b.first.first / len) {
    return a.first.second < b.first.second;
  }
  return a.first.first < b.first.first;
}
long long poww(long long a, long long step) {
  if (step == 0) {
    return 1;
  }
  if (step % 2) {
    return (a * poww(a, step - 1) % (int)(1e9 + 7));
  } else {
    long long buf = poww(a, step / 2);
    return (buf * buf) % (int)(1e9 + 7);
  }
}
vector<int> fenvik;
int n;
int maxV = -1;
vector<int> isOt;
vector<vector<int> > g, rg;
vector<int> flag;
vector<int> ts;
vector<int> path;
bool isOtv = true;
void dfs(int v, int len = 0) {
  flag[v] = 1;
  if (len > maxV) {
    maxV = max(len, maxV);
    path.clear();
    isOtv = false;
  }
  for (long long i = 0; i < g[v].size(); i++) {
    if (!flag[g[v][i]]) dfs(g[v][i], len + 1);
  }
  if (!isOtv) {
    path.push_back(v);
  }
}
int main() {
  int n;
  cin >> n;
  isOt.resize(n);
  g.resize(n);
  vector<int> start;
  flag.resize(n, 0);
  vector<int> otel;
  for (long long i = 0; i < n; i++) {
    cin >> isOt[i];
    if (isOt[i]) {
      otel.push_back(i);
    }
  }
  vector<int> cp(n, 0);
  for (long long i = 0; i < n; i++) {
    int buf;
    cin >> buf;
    if (buf) {
      buf--;
      cp[buf]++;
      if (cp[buf] > 1) flag[buf] = 1;
      g[i].push_back(buf);
    } else {
      start.push_back(i);
    }
  }
  flag.resize(n, 0);
  vector<int> res;
  for (long long i = 0; i < otel.size(); i++) {
    isOtv = true;
    path.clear();
    int oldV = maxV;
    dfs(otel[i]);
    if (oldV < maxV) {
      res = path;
      oldV = maxV;
    }
  }
  cout << res.size() << endl;
  for (long long i = 0; i < res.size(); i++) {
    cout << res[i] + 1 << ' ';
  }
  return 0;
}

```