#include <bits/stdc++.h>
int t[100010], a[100010];
int dig[100010];
int step[100010], vis[100010];
int n;
void dfs(int x) {
  vis[x] = 1;
  if (a[x] == 0 || t[a[x]] == 1 || dig[a[x]] != 1 || vis[a[x]] == 1) {
    step[x] = 1;
    return;
  }
  if (vis[a[x]] == 0) dfs(a[x]);
  step[x] = step[a[x]] + 1;
}
void output(int x) {
  int pre = a[x];
  vis[x] = 1;
  if (pre == 0 || t[pre] == 1 || dig[pre] != 1 || vis[pre] == 1)
    ;
  else
    output(pre);
  printf("%d ", x);
}
int main() {
  int i, ans;
  while (~scanf("%d", &n)) {
    memset(dig, 0, sizeof(dig));
    for (i = 1; i <= n; i++) scanf("%d", &t[i]);
    for (i = 1; i <= n; i++) scanf("%d", &a[i]), dig[a[i]]++;
    ans = -1;
    memset(vis, 0, sizeof(vis));
    for (i = 1; i <= n; i++)
      if (t[i]) {
        dfs(i);
        if (ans == -1 || step[ans] < step[i]) ans = i;
      }
    printf("%d\n", step[ans]);
    memset(vis, 0, sizeof(vis));
    output(ans);
    printf("\n");
  }
  return 0;
}

### Prompt

Generate 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>
int t[100010], a[100010];
int dig[100010];
int step[100010], vis[100010];
int n;
void dfs(int x) {
  vis[x] = 1;
  if (a[x] == 0 || t[a[x]] == 1 || dig[a[x]] != 1 || vis[a[x]] == 1) {
    step[x] = 1;
    return;
  }
  if (vis[a[x]] == 0) dfs(a[x]);
  step[x] = step[a[x]] + 1;
}
void output(int x) {
  int pre = a[x];
  vis[x] = 1;
  if (pre == 0 || t[pre] == 1 || dig[pre] != 1 || vis[pre] == 1)
    ;
  else
    output(pre);
  printf("%d ", x);
}
int main() {
  int i, ans;
  while (~scanf("%d", &n)) {
    memset(dig, 0, sizeof(dig));
    for (i = 1; i <= n; i++) scanf("%d", &t[i]);
    for (i = 1; i <= n; i++) scanf("%d", &a[i]), dig[a[i]]++;
    ans = -1;
    memset(vis, 0, sizeof(vis));
    for (i = 1; i <= n; i++)
      if (t[i]) {
        dfs(i);
        if (ans == -1 || step[ans] < step[i]) ans = i;
      }
    printf("%d\n", step[ans]);
    memset(vis, 0, sizeof(vis));
    output(ans);
    printf("\n");
  }
  return 0;
}

```