#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 100 * 1000 + 1;
bool obj[MAX_N];
vector<int> graph[MAX_N];
int N;
int dad[MAX_N];
int dp[MAX_N];
bool mark[MAX_N];
queue<int> BB;
int dad2[MAX_N];
int dad3[MAX_N];
void input() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> obj[i];
  for (int i = 0; i < N; i++) {
    int a;
    cin >> a;
    if (a) {
      graph[a - 1].push_back(i);
      dad[i] = a - 1;
    } else
      dad[i] = -1;
  }
}
int run() {
  for (int i = 0; i < N; i++) {
    if (graph[i].size() <= 1) mark[i] = true;
  }
  for (int i = 0; i < N; i++) {
    if (obj[i]) BB.push(i);
    dad2[i] = -1;
    dad3[i] = 0;
  }
  while (BB.size()) {
    int y = BB.front();
    if (mark[dad[y]] && dad[y] != -1) {
      dad2[dad[y]] = y;
      dad3[dad[y]] = dad3[y] + 1;
      BB.push(dad[y]);
    }
    BB.pop();
  }
  int y = 0;
  for (int i = 1; i < N; i++) {
    if (dad3[y] < dad3[i]) y = i;
  }
  if (y == 0 && dad3[y] == 0) {
    for (int i = 0; i < N; i++)
      if (obj[i]) {
        y = i;
        break;
      }
  }
  return y;
}
void output(int ans) {
  cout << dad3[ans] + 1 << endl;
  int y = ans;
  while (y != -1) {
    cout << y + 1 << " ";
    y = dad2[y];
  }
  cout << endl;
}
int main() {
  input();
  output(run());
}

### Prompt

Please create a solution in CPP 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 MAX_N = 100 * 1000 + 1;
bool obj[MAX_N];
vector<int> graph[MAX_N];
int N;
int dad[MAX_N];
int dp[MAX_N];
bool mark[MAX_N];
queue<int> BB;
int dad2[MAX_N];
int dad3[MAX_N];
void input() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> obj[i];
  for (int i = 0; i < N; i++) {
    int a;
    cin >> a;
    if (a) {
      graph[a - 1].push_back(i);
      dad[i] = a - 1;
    } else
      dad[i] = -1;
  }
}
int run() {
  for (int i = 0; i < N; i++) {
    if (graph[i].size() <= 1) mark[i] = true;
  }
  for (int i = 0; i < N; i++) {
    if (obj[i]) BB.push(i);
    dad2[i] = -1;
    dad3[i] = 0;
  }
  while (BB.size()) {
    int y = BB.front();
    if (mark[dad[y]] && dad[y] != -1) {
      dad2[dad[y]] = y;
      dad3[dad[y]] = dad3[y] + 1;
      BB.push(dad[y]);
    }
    BB.pop();
  }
  int y = 0;
  for (int i = 1; i < N; i++) {
    if (dad3[y] < dad3[i]) y = i;
  }
  if (y == 0 && dad3[y] == 0) {
    for (int i = 0; i < N; i++)
      if (obj[i]) {
        y = i;
        break;
      }
  }
  return y;
}
void output(int ans) {
  cout << dad3[ans] + 1 << endl;
  int y = ans;
  while (y != -1) {
    cout << y + 1 << " ";
    y = dad2[y];
  }
  cout << endl;
}
int main() {
  input();
  output(run());
}

```