#include <bits/stdc++.h>
using namespace std;
const int Mod = 998244353;
int dcmp(double x) {
  if (fabs(x) <= 1e-9) return 0;
  return x < 0 ? -1 : 1;
}
struct Point {
  double x, y;
  double ang;
  Point() {}
  Point(double x, double y) : x(x), y(y) {}
  Point operator-(const Point &a) const { return Point(x - a.x, y - a.y); }
  Point operator+(const Point &a) const { return Point(x + a.x, y + a.y); }
  Point operator*(const double &a) const { return Point(x * a, y * a); }
  bool operator<(const Point &a) const {
    if (dcmp(ang) == 0) return (x * x + y * y) <= a.x * a.x + a.y * a.y;
    return ang < a.ang;
  }
  bool operator==(const Point &a) const { return x == a.x && y == a.y; }
};
struct cycle {
  Point p;
  long long r;
  cycle() {}
  cycle(Point p, long long r) : p(p), r(r) {}
};
Point Rotate(Point v, double ang) {
  return Point(v.x * cos(ang) - v.y * sin(ang),
               v.x * sin(ang) + v.y * cos(ang));
}
double Dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double dis(Point a) { return sqrt(Dot(a, a)); }
double Cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
int f[200005], s[200005];
int dp[200005], v, n, cut[200005];
int dfs(int now) {
  if (!now) return 0;
  if (f[now] == now) return 0;
  if (dp[now] != -1) return dp[now];
  if (s[now]) return dp[now] = 0;
  if (cut[now] != 1) return dp[now] = 0;
  return dp[now] = dfs(f[now]) + 1;
}
void rdfs(int now) {
  if (!now) return;
  if (s[now]) return;
  if (cut[now] != 1) return;
  if (f[now]) rdfs(f[now]);
  cout << now << " ";
}
void solve() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  memset(dp, -1, sizeof(dp));
  memset(cut, 0, sizeof(cut));
  for (int i = 1; i <= n; i++) cin >> s[i];
  for (int i = 1; i <= n; i++) {
    cin >> v;
    f[i] = v, cut[v]++;
  }
  int ans = 0, id = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i]) {
      int cnt = dfs(f[i]) + 1;
      if (cnt >= ans) ans = cnt, id = i;
    }
  }
  cout << ans << endl;
  rdfs(f[id]);
  cout << id << endl;
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) solve();
  return 0;
}

### 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;
const int Mod = 998244353;
int dcmp(double x) {
  if (fabs(x) <= 1e-9) return 0;
  return x < 0 ? -1 : 1;
}
struct Point {
  double x, y;
  double ang;
  Point() {}
  Point(double x, double y) : x(x), y(y) {}
  Point operator-(const Point &a) const { return Point(x - a.x, y - a.y); }
  Point operator+(const Point &a) const { return Point(x + a.x, y + a.y); }
  Point operator*(const double &a) const { return Point(x * a, y * a); }
  bool operator<(const Point &a) const {
    if (dcmp(ang) == 0) return (x * x + y * y) <= a.x * a.x + a.y * a.y;
    return ang < a.ang;
  }
  bool operator==(const Point &a) const { return x == a.x && y == a.y; }
};
struct cycle {
  Point p;
  long long r;
  cycle() {}
  cycle(Point p, long long r) : p(p), r(r) {}
};
Point Rotate(Point v, double ang) {
  return Point(v.x * cos(ang) - v.y * sin(ang),
               v.x * sin(ang) + v.y * cos(ang));
}
double Dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double dis(Point a) { return sqrt(Dot(a, a)); }
double Cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
int f[200005], s[200005];
int dp[200005], v, n, cut[200005];
int dfs(int now) {
  if (!now) return 0;
  if (f[now] == now) return 0;
  if (dp[now] != -1) return dp[now];
  if (s[now]) return dp[now] = 0;
  if (cut[now] != 1) return dp[now] = 0;
  return dp[now] = dfs(f[now]) + 1;
}
void rdfs(int now) {
  if (!now) return;
  if (s[now]) return;
  if (cut[now] != 1) return;
  if (f[now]) rdfs(f[now]);
  cout << now << " ";
}
void solve() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  memset(dp, -1, sizeof(dp));
  memset(cut, 0, sizeof(cut));
  for (int i = 1; i <= n; i++) cin >> s[i];
  for (int i = 1; i <= n; i++) {
    cin >> v;
    f[i] = v, cut[v]++;
  }
  int ans = 0, id = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i]) {
      int cnt = dfs(f[i]) + 1;
      if (cnt >= ans) ans = cnt, id = i;
    }
  }
  cout << ans << endl;
  rdfs(f[id]);
  cout << id << endl;
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) solve();
  return 0;
}

```