#include <bits/stdc++.h>
using namespace std;
double PI = acos(-1);
double EPS = 1e-7;
int INF = 1000000000;
int MOD = 1000000007;
int MAXINT = 2147483647;
long long INFLL = 1000000000000000000LL;
long long MAXLL = 9223372036854775807LL;
int ty[100005];
int pr[100005];
int sz[100005];
int vs[100005];
vector<int> lol;
int go(int k) {
  if (vs[k] != -1) return vs[k];
  if (sz[pr[k]] == 1 && pr[k]) {
    return vs[k] = 1 + go(pr[k]);
  } else
    return vs[k] = 0;
}
vector<int> ans;
void track(int k) {
  if (sz[pr[k]] == 1 && pr[k]) {
    track(pr[k]);
  }
  ans.push_back(k);
}
int main() {
  memset(vs, -1, sizeof(vs));
  int n;
  scanf("%d", &n);
  for (int(a) = (1); (a) <= (n); (a)++) {
    scanf("%d", &ty[a]);
  }
  for (int(a) = (1); (a) <= (n); (a)++) {
    scanf("%d", &pr[a]);
    sz[pr[a]]++;
  }
  pair<int, int> maks = make_pair(0, 0);
  for (int(a) = (1); (a) <= (n); (a)++) {
    if (ty[a]) (maks) = max((maks), (make_pair(1 + go(a), a)));
  }
  track(maks.second);
  printf("%d\n", (int)ans.size());
  for (int(a) = (0); (a) <= ((int)ans.size() - 1); (a)++) {
    printf("%d ", ans[a]);
  }
  puts("");
}

### 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;
double PI = acos(-1);
double EPS = 1e-7;
int INF = 1000000000;
int MOD = 1000000007;
int MAXINT = 2147483647;
long long INFLL = 1000000000000000000LL;
long long MAXLL = 9223372036854775807LL;
int ty[100005];
int pr[100005];
int sz[100005];
int vs[100005];
vector<int> lol;
int go(int k) {
  if (vs[k] != -1) return vs[k];
  if (sz[pr[k]] == 1 && pr[k]) {
    return vs[k] = 1 + go(pr[k]);
  } else
    return vs[k] = 0;
}
vector<int> ans;
void track(int k) {
  if (sz[pr[k]] == 1 && pr[k]) {
    track(pr[k]);
  }
  ans.push_back(k);
}
int main() {
  memset(vs, -1, sizeof(vs));
  int n;
  scanf("%d", &n);
  for (int(a) = (1); (a) <= (n); (a)++) {
    scanf("%d", &ty[a]);
  }
  for (int(a) = (1); (a) <= (n); (a)++) {
    scanf("%d", &pr[a]);
    sz[pr[a]]++;
  }
  pair<int, int> maks = make_pair(0, 0);
  for (int(a) = (1); (a) <= (n); (a)++) {
    if (ty[a]) (maks) = max((maks), (make_pair(1 + go(a), a)));
  }
  track(maks.second);
  printf("%d\n", (int)ans.size());
  for (int(a) = (0); (a) <= ((int)ans.size() - 1); (a)++) {
    printf("%d ", ans[a]);
  }
  puts("");
}

```