#include <bits/stdc++.h>
using namespace std;
int type[100100 + 1], p[100100 + 1], deg[100100 + 1], dist[100100 + 1];
queue<int> q;
void traverse(int s, int u, int cnt) {
  if (type[u] == 1 || deg[u] != 1 || u == -1) {
    dist[s] = cnt - 1;
    return;
  }
  traverse(s, p[u], cnt + 1);
}
void print_path(int u, int dist) {
  if (dist == 0) {
    q.push(u);
    return;
  }
  print_path(p[u], dist - 1);
  q.push(u);
}
int main(void) {
  int N, u, root = 1;
  scanf("%d", &N);
  vector<int> h;
  for (int i = 1; i <= N; i++) {
    p[i] = -1;
    scanf("%d", &type[i]);
    if (type[i]) {
      h.push_back(i);
    }
  }
  for (int i = 1; i <= N; i++) {
    scanf("%d", &u);
    if (u) {
      p[i] = u;
      deg[u]++;
    }
  }
  for (int i = 0; i < h.size(); i++) {
    traverse(h[i], p[h[i]], 2);
    if (dist[h[i]] >= dist[root]) {
      root = h[i];
    }
  }
  printf("%d\n", dist[root]);
  print_path(root, dist[root] - 1);
  printf("%d", q.front());
  q.pop();
  while (!q.empty()) {
    int u = q.front();
    printf(" %d", u);
    q.pop();
  }
  printf("\n");
  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;
int type[100100 + 1], p[100100 + 1], deg[100100 + 1], dist[100100 + 1];
queue<int> q;
void traverse(int s, int u, int cnt) {
  if (type[u] == 1 || deg[u] != 1 || u == -1) {
    dist[s] = cnt - 1;
    return;
  }
  traverse(s, p[u], cnt + 1);
}
void print_path(int u, int dist) {
  if (dist == 0) {
    q.push(u);
    return;
  }
  print_path(p[u], dist - 1);
  q.push(u);
}
int main(void) {
  int N, u, root = 1;
  scanf("%d", &N);
  vector<int> h;
  for (int i = 1; i <= N; i++) {
    p[i] = -1;
    scanf("%d", &type[i]);
    if (type[i]) {
      h.push_back(i);
    }
  }
  for (int i = 1; i <= N; i++) {
    scanf("%d", &u);
    if (u) {
      p[i] = u;
      deg[u]++;
    }
  }
  for (int i = 0; i < h.size(); i++) {
    traverse(h[i], p[h[i]], 2);
    if (dist[h[i]] >= dist[root]) {
      root = h[i];
    }
  }
  printf("%d\n", dist[root]);
  print_path(root, dist[root] - 1);
  printf("%d", q.front());
  q.pop();
  while (!q.empty()) {
    int u = q.front();
    printf(" %d", u);
    q.pop();
  }
  printf("\n");
  return 0;
}

```