#include <bits/stdc++.h>
using namespace std;
int Max(int a, int b) { return (((b - a) >> (32 - 1)) & (a ^ b)) ^ b; }
int Min(int a, int b) { return (((b - a) >> (32 - 1)) & (a ^ b)) ^ a; }
long long GCD(long long a, long long b) { return b != 0 ? GCD(b, a % b) : a; }
long long LCM(long long a, long long b) { return a * (b / GCD(a, b)); }
long long Pow(long long n, long long e) {
  if (e == 0) return 1;
  if (e == 1)
    return n;
  else if (e & 1) {
    long long t = Pow(n, e / 2);
    return n * t * t;
  } else {
    long long t = Pow(n, e / 2);
    return t * t;
  }
}
const int MAX = 100005;
int t[MAX];
vector<int> edge[MAX];
int dp[MAX], nextt[MAX];
bool vis[MAX];
int DFS(int id) {
  if (t[id] == 1) return 1;
  if (vis[id]) return -100000000;
  if (dp[id] != -1) return dp[id];
  vis[id] = true;
  int res = -100000000, m = 0;
  for (int i = 0, max_ = ((int)(edge[id]).size()); i < max_; ++i) {
    int nid = edge[id][i];
    if (t[nid] == 0) ++m;
    int r = 1 + DFS(nid);
    if (r > res) {
      res = r;
      nextt[id] = nid;
    }
  }
  if ((int)(edge[id]).size() > 1)
    return dp[id] = -100000000;
  else
    return dp[id] = res;
}
void Print(int id) {
  printf("%d ", id + 1);
  if (nextt[id] >= 0) Print(nextt[id]);
}
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 0, max_ = (N); i < max_; ++i) scanf("%d", &t[i]);
  for (int i = 0, max_ = (N); i < max_; ++i) {
    int a;
    scanf("%d", &a);
    if (a > 0) {
      --a;
      edge[a].push_back(i);
    }
  }
  memset(dp, -1, sizeof(dp));
  ;
  memset(nextt, -1, sizeof(nextt));
  ;
  int res = 0, resID = -1;
  for (int i = 0, max_ = (N); i < max_; ++i) {
    memset(vis, false, sizeof(vis));
    ;
    int r = DFS(i);
    if (r > res) {
      res = r;
      resID = i;
    }
  }
  printf("%d\n", res);
  Print(resID);
  return 0;
}

### Prompt

Develop a solution in cpp 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 Max(int a, int b) { return (((b - a) >> (32 - 1)) & (a ^ b)) ^ b; }
int Min(int a, int b) { return (((b - a) >> (32 - 1)) & (a ^ b)) ^ a; }
long long GCD(long long a, long long b) { return b != 0 ? GCD(b, a % b) : a; }
long long LCM(long long a, long long b) { return a * (b / GCD(a, b)); }
long long Pow(long long n, long long e) {
  if (e == 0) return 1;
  if (e == 1)
    return n;
  else if (e & 1) {
    long long t = Pow(n, e / 2);
    return n * t * t;
  } else {
    long long t = Pow(n, e / 2);
    return t * t;
  }
}
const int MAX = 100005;
int t[MAX];
vector<int> edge[MAX];
int dp[MAX], nextt[MAX];
bool vis[MAX];
int DFS(int id) {
  if (t[id] == 1) return 1;
  if (vis[id]) return -100000000;
  if (dp[id] != -1) return dp[id];
  vis[id] = true;
  int res = -100000000, m = 0;
  for (int i = 0, max_ = ((int)(edge[id]).size()); i < max_; ++i) {
    int nid = edge[id][i];
    if (t[nid] == 0) ++m;
    int r = 1 + DFS(nid);
    if (r > res) {
      res = r;
      nextt[id] = nid;
    }
  }
  if ((int)(edge[id]).size() > 1)
    return dp[id] = -100000000;
  else
    return dp[id] = res;
}
void Print(int id) {
  printf("%d ", id + 1);
  if (nextt[id] >= 0) Print(nextt[id]);
}
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 0, max_ = (N); i < max_; ++i) scanf("%d", &t[i]);
  for (int i = 0, max_ = (N); i < max_; ++i) {
    int a;
    scanf("%d", &a);
    if (a > 0) {
      --a;
      edge[a].push_back(i);
    }
  }
  memset(dp, -1, sizeof(dp));
  ;
  memset(nextt, -1, sizeof(nextt));
  ;
  int res = 0, resID = -1;
  for (int i = 0, max_ = (N); i < max_; ++i) {
    memset(vis, false, sizeof(vis));
    ;
    int r = DFS(i);
    if (r > res) {
      res = r;
      resID = i;
    }
  }
  printf("%d\n", res);
  Print(resID);
  return 0;
}

```