#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const long long INF = 0x3f3f3f3f3f3f;
const int mod = 1000000007;
int n, x, num, po, flag;
int chu[maxn], a[maxn];
struct node {
  int to, next;
} edge[maxn];
int head[maxn], cnt;
int vis[maxn], pre[maxn];
vector<int> vec;
int pos = 0, ans = -1;
void addEdge(int u, int v) {
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}
void dfs(int u, int cen) {
  if (cen > ans) {
    ans = cen;
    pos = u;
  }
  int v = pre[u];
  if (vis[v] == 0 && chu[v] == 1 && v) dfs(v, cen + 1);
}
void dfs2(int u) {
  vec.push_back(u);
  int i = head[u];
  if (i == -1) return;
  if (a[u] == 1) {
    return;
  }
  dfs2(edge[i].to);
}
int main() {
  scanf("%d", &n);
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) vis[i] = 1;
  }
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &pre[i]);
    if (pre[i]) {
      addEdge(pre[i], i);
      chu[pre[i]]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) {
      dfs(i, 1);
    }
  }
  printf("%d\n", ans);
  dfs2(pos);
  for (int i = 0; i < vec.size(); i++) {
    printf("%d ", vec[i]);
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const int maxn = 2e5 + 10;
const long long INF = 0x3f3f3f3f3f3f;
const int mod = 1000000007;
int n, x, num, po, flag;
int chu[maxn], a[maxn];
struct node {
  int to, next;
} edge[maxn];
int head[maxn], cnt;
int vis[maxn], pre[maxn];
vector<int> vec;
int pos = 0, ans = -1;
void addEdge(int u, int v) {
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}
void dfs(int u, int cen) {
  if (cen > ans) {
    ans = cen;
    pos = u;
  }
  int v = pre[u];
  if (vis[v] == 0 && chu[v] == 1 && v) dfs(v, cen + 1);
}
void dfs2(int u) {
  vec.push_back(u);
  int i = head[u];
  if (i == -1) return;
  if (a[u] == 1) {
    return;
  }
  dfs2(edge[i].to);
}
int main() {
  scanf("%d", &n);
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) vis[i] = 1;
  }
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &pre[i]);
    if (pre[i]) {
      addEdge(pre[i], i);
      chu[pre[i]]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) {
      dfs(i, 1);
    }
  }
  printf("%d\n", ans);
  dfs2(pos);
  for (int i = 0; i < vec.size(); i++) {
    printf("%d ", vec[i]);
  }
  return 0;
}

```