#include <bits/stdc++.h>
using namespace std;
int n;
int hot[100005], du[100005];
int pre[100005];
int len[100005];
int t;
int p[100005];
void init() {
  for (int i = 0; i < 100005; i++) pre[i] = i;
  memset(hot, 0, sizeof(hot));
  memset(du, 0, sizeof(du));
  memset(len, 0, sizeof(len));
}
int c;
int find(int x, int s) {
  if (x == pre[x]) {
    if (du[x] == 1)
      return s + 1;
    else
      return s;
  } else {
    if (du[x] == 1)
      return find(pre[x], s + 1);
    else
      return s;
  }
}
void print(int x) {
  if (du[x] == 1) {
    p[t++] = x;
    if (x == pre[x]) return;
    print(pre[x]);
  }
}
void solve() {
  for (int i = 1; i <= n; i++) {
    if (hot[i]) {
      len[i] = find(pre[i], 1);
    }
  }
  int max = -1, k = 0;
  for (int i = 1; i <= n; i++) {
    if (len[i] > max) {
      max = len[i];
      k = i;
    }
  }
  cout << len[k] << endl;
  t = 0;
  p[t++] = k;
  print(pre[k]);
  for (int i = t - 1; i >= 0; i--) cout << p[i] << " ";
  cout << endl;
}
void input() {
  while (cin >> n) {
    init();
    for (int i = 1; i <= n; i++) {
      cin >> c;
      hot[i] = c;
    }
    for (int i = 1; i <= n; i++) {
      cin >> c;
      if (c == 0)
        ;
      else {
        du[c]++;
        pre[i] = c;
      }
    }
    solve();
  }
}
int main() {
  input();
  return 0;
}

### Prompt

In CPP, your task is to solve 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 n;
int hot[100005], du[100005];
int pre[100005];
int len[100005];
int t;
int p[100005];
void init() {
  for (int i = 0; i < 100005; i++) pre[i] = i;
  memset(hot, 0, sizeof(hot));
  memset(du, 0, sizeof(du));
  memset(len, 0, sizeof(len));
}
int c;
int find(int x, int s) {
  if (x == pre[x]) {
    if (du[x] == 1)
      return s + 1;
    else
      return s;
  } else {
    if (du[x] == 1)
      return find(pre[x], s + 1);
    else
      return s;
  }
}
void print(int x) {
  if (du[x] == 1) {
    p[t++] = x;
    if (x == pre[x]) return;
    print(pre[x]);
  }
}
void solve() {
  for (int i = 1; i <= n; i++) {
    if (hot[i]) {
      len[i] = find(pre[i], 1);
    }
  }
  int max = -1, k = 0;
  for (int i = 1; i <= n; i++) {
    if (len[i] > max) {
      max = len[i];
      k = i;
    }
  }
  cout << len[k] << endl;
  t = 0;
  p[t++] = k;
  print(pre[k]);
  for (int i = t - 1; i >= 0; i--) cout << p[i] << " ";
  cout << endl;
}
void input() {
  while (cin >> n) {
    init();
    for (int i = 1; i <= n; i++) {
      cin >> c;
      hot[i] = c;
    }
    for (int i = 1; i <= n; i++) {
      cin >> c;
      if (c == 0)
        ;
      else {
        du[c]++;
        pre[i] = c;
      }
    }
    solve();
  }
}
int main() {
  input();
  return 0;
}

```