#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int Max = 1e5 + 10;
int n, s;
int type[Max], d[Max];
vector<int> edge[Max];
stack<int> path[Max];
bool vis[Max];
void dfs() {
  for (int i = 1; i <= n; i++) {
    if (type[i] == 1) {
      path[i].push(i);
      if (edge[i].size() == 0) continue;
      int now = edge[i][0];
      while (true) {
        if (d[now] > 1 || type[now] == 1) {
          break;
        }
        path[i].push(now);
        if (edge[now].size() == 0) break;
        now = edge[now][0];
      }
    }
  }
  int dis = 0;
  int id = -1;
  for (int i = 1; i <= n; i++) {
    if (type[i] == 1) {
      if (dis < path[i].size()) {
        dis = path[i].size();
        id = i;
      }
    }
  }
  printf("%d\n", dis);
  while (!path[id].empty()) {
    printf("%d ", path[id].top());
    path[id].pop();
  }
  printf("\n");
}
int main() {
  while (scanf("%d", &n) != EOF) {
    s = -1;
    for (int i = 1; i <= n; i++) {
      scanf("%d", type + i);
      if (type[i] == 1 && s == -1) s = i;
      while (!path[i].empty()) path[i].pop();
      d[i] = 0;
      edge[i].clear();
    }
    for (int v = 1; v <= n; v++) {
      int u;
      scanf("%d", &u);
      if (u != 0) {
        edge[v].push_back(u);
        d[u]++;
      }
    }
    dfs();
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int Max = 1e5 + 10;
int n, s;
int type[Max], d[Max];
vector<int> edge[Max];
stack<int> path[Max];
bool vis[Max];
void dfs() {
  for (int i = 1; i <= n; i++) {
    if (type[i] == 1) {
      path[i].push(i);
      if (edge[i].size() == 0) continue;
      int now = edge[i][0];
      while (true) {
        if (d[now] > 1 || type[now] == 1) {
          break;
        }
        path[i].push(now);
        if (edge[now].size() == 0) break;
        now = edge[now][0];
      }
    }
  }
  int dis = 0;
  int id = -1;
  for (int i = 1; i <= n; i++) {
    if (type[i] == 1) {
      if (dis < path[i].size()) {
        dis = path[i].size();
        id = i;
      }
    }
  }
  printf("%d\n", dis);
  while (!path[id].empty()) {
    printf("%d ", path[id].top());
    path[id].pop();
  }
  printf("\n");
}
int main() {
  while (scanf("%d", &n) != EOF) {
    s = -1;
    for (int i = 1; i <= n; i++) {
      scanf("%d", type + i);
      if (type[i] == 1 && s == -1) s = i;
      while (!path[i].empty()) path[i].pop();
      d[i] = 0;
      edge[i].clear();
    }
    for (int v = 1; v <= n; v++) {
      int u;
      scanf("%d", &u);
      if (u != 0) {
        edge[v].push_back(u);
        d[u]++;
      }
    }
    dfs();
  }
  return 0;
}

```