#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int n;
int type[maxn], dig[maxn], fa[maxn], pre[maxn];
int head[maxn], ecnt;
stack<int> path;
struct eed {
  int next, to;
} edge[maxn];
void addedge(int u, int v) {
  edge[ecnt].next = head[u];
  edge[ecnt].to = v;
  head[u] = ecnt++;
}
int ans = 0, goal;
void dfs(int u, int len) {
  if (type[u]) {
    if (len > ans) {
      ans = len;
      goal = u;
    }
    return;
  }
  if (dig[u] != 1) return;
  pre[edge[head[u]].to] = u;
  dfs(edge[head[u]].to, len + 1);
}
int main() {
  scanf("%d", &n);
  int i;
  for (i = 1; i <= n; ++i) scanf("%d", type + i);
  memset(dig, 0, sizeof(dig));
  memset(head, -1, sizeof(head));
  ecnt = 0;
  for (i = 1; i <= n; ++i) {
    scanf("%d", fa + i);
    if (fa[i]) {
      addedge(fa[i], i);
      dig[fa[i]]++;
    }
  }
  memset(pre, -1, sizeof(pre));
  for (i = 1; i <= n; ++i)
    if (!fa[i] || type[fa[i]] || dig[fa[i]] != 1) dfs(i, 1);
  printf("%d\n", ans);
  for (i = goal; i != -1; i = pre[i]) path.push(i);
  while (!path.empty()) {
    printf("%d ", path.top());
    path.pop();
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
const int maxn = 1e5 + 5;
int n;
int type[maxn], dig[maxn], fa[maxn], pre[maxn];
int head[maxn], ecnt;
stack<int> path;
struct eed {
  int next, to;
} edge[maxn];
void addedge(int u, int v) {
  edge[ecnt].next = head[u];
  edge[ecnt].to = v;
  head[u] = ecnt++;
}
int ans = 0, goal;
void dfs(int u, int len) {
  if (type[u]) {
    if (len > ans) {
      ans = len;
      goal = u;
    }
    return;
  }
  if (dig[u] != 1) return;
  pre[edge[head[u]].to] = u;
  dfs(edge[head[u]].to, len + 1);
}
int main() {
  scanf("%d", &n);
  int i;
  for (i = 1; i <= n; ++i) scanf("%d", type + i);
  memset(dig, 0, sizeof(dig));
  memset(head, -1, sizeof(head));
  ecnt = 0;
  for (i = 1; i <= n; ++i) {
    scanf("%d", fa + i);
    if (fa[i]) {
      addedge(fa[i], i);
      dig[fa[i]]++;
    }
  }
  memset(pre, -1, sizeof(pre));
  for (i = 1; i <= n; ++i)
    if (!fa[i] || type[fa[i]] || dig[fa[i]] != 1) dfs(i, 1);
  printf("%d\n", ans);
  for (i = goal; i != -1; i = pre[i]) path.push(i);
  while (!path.empty()) {
    printf("%d ", path.top());
    path.pop();
  }
  return 0;
}

```