#include <bits/stdc++.h>
using namespace std;
vector<long long> v[123456];
long long rev[123456];
bool ishotel[123456];
long long N;
vector<long long> start;
vector<long long> ans;
vector<long long> temp;
long long vis[123456];
void dfs(long long node, long long now) {
  vis[node] = now;
  long long i, j;
  if (v[node].size() == 1 && rev[node] != start[now - 1]) {
    temp.push_back(node);
    if (rev[node] != 0) dfs(rev[node], now);
  }
}
int main() {
  long long i, j;
  scanf("%lld", &N);
  for (i = 1; i <= N; i++) {
    scanf("%lld", &ishotel[i]);
    if (ishotel[i]) start.push_back(i);
  }
  for (i = 1; i <= N; i++) {
    long long X;
    scanf("%lld", &X);
    if (X != 0) {
      v[X].push_back(i);
      rev[i] = (X);
    }
  }
  memset(vis, 0, sizeof(vis));
  for (i = 0; i < start.size(); i++) {
    temp.clear();
    temp.push_back(start[i]);
    dfs(rev[start[i]], i + 1);
    if (temp.size() > ans.size()) ans = temp;
  }
  printf("%lld\n", (long long)ans.size());
  for (i = ans.size() - 1; i >= 0; i--) printf("%lld ", ans[i]);
  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;
vector<long long> v[123456];
long long rev[123456];
bool ishotel[123456];
long long N;
vector<long long> start;
vector<long long> ans;
vector<long long> temp;
long long vis[123456];
void dfs(long long node, long long now) {
  vis[node] = now;
  long long i, j;
  if (v[node].size() == 1 && rev[node] != start[now - 1]) {
    temp.push_back(node);
    if (rev[node] != 0) dfs(rev[node], now);
  }
}
int main() {
  long long i, j;
  scanf("%lld", &N);
  for (i = 1; i <= N; i++) {
    scanf("%lld", &ishotel[i]);
    if (ishotel[i]) start.push_back(i);
  }
  for (i = 1; i <= N; i++) {
    long long X;
    scanf("%lld", &X);
    if (X != 0) {
      v[X].push_back(i);
      rev[i] = (X);
    }
  }
  memset(vis, 0, sizeof(vis));
  for (i = 0; i < start.size(); i++) {
    temp.clear();
    temp.push_back(start[i]);
    dfs(rev[start[i]], i + 1);
    if (temp.size() > ans.size()) ans = temp;
  }
  printf("%lld\n", (long long)ans.size());
  for (i = ans.size() - 1; i >= 0; i--) printf("%lld ", ans[i]);
  return 0;
}

```