#include <bits/stdc++.h>
const int maxn = 100000 + 5;
struct PP {
  int to, next;
} edge[maxn * 5];
int head[maxn], vis[maxn], ru[maxn], pa[maxn], path[maxn], is[maxn], a[maxn],
    cnt[maxn];
int E, dep[maxn];
void init() {
  memset(head, -1, sizeof(head));
  E = 0;
}
void newedge(int u, int to) {
  edge[E].to = to;
  edge[E].next = head[u];
  head[u] = E++;
}
int max(int a, int b) { return a > b ? a : b; }
int mx, ans, flag;
void dfs(int u) {
  mx = max(mx, dep[u]);
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int to = edge[i].to;
    dep[to] = dep[u] + 1;
    dfs(to);
  }
}
void DFS(int u, int d) {
  if (flag) return;
  if (d == ans + 1) {
    for (int i = ans; i >= 1; i--) printf("%d ", path[i]);
    puts("");
    flag = 1;
    return;
  }
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int to = edge[i].to;
    path[d] = to;
    DFS(to, d + 1);
    if (flag) return;
  }
}
int main() {
  init();
  int n, u;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &is[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
  }
  for (int i = 1; i <= n; i++)
    if (!is[i] && a[i]) {
      if (cnt[i] == 1 && cnt[a[i]] == 1 && !is[a[i]]) newedge(i, a[i]);
    } else if (a[i] && cnt[a[i]] == 1)
      newedge(i, a[i]);
  for (int i = 1; i <= n; i++) cnt[i] = 0;
  ans = 0;
  int id;
  for (int i = 1; i <= n; i++)
    if (is[i]) {
      mx = 0;
      dep[i] = 1;
      dfs(i);
      if (mx > ans) id = i, ans = mx;
    }
  printf("%d\n", ans);
  path[1] = id;
  flag = 0;
  DFS(id, 2);
}

### Prompt

Please formulate 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>
const int maxn = 100000 + 5;
struct PP {
  int to, next;
} edge[maxn * 5];
int head[maxn], vis[maxn], ru[maxn], pa[maxn], path[maxn], is[maxn], a[maxn],
    cnt[maxn];
int E, dep[maxn];
void init() {
  memset(head, -1, sizeof(head));
  E = 0;
}
void newedge(int u, int to) {
  edge[E].to = to;
  edge[E].next = head[u];
  head[u] = E++;
}
int max(int a, int b) { return a > b ? a : b; }
int mx, ans, flag;
void dfs(int u) {
  mx = max(mx, dep[u]);
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int to = edge[i].to;
    dep[to] = dep[u] + 1;
    dfs(to);
  }
}
void DFS(int u, int d) {
  if (flag) return;
  if (d == ans + 1) {
    for (int i = ans; i >= 1; i--) printf("%d ", path[i]);
    puts("");
    flag = 1;
    return;
  }
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int to = edge[i].to;
    path[d] = to;
    DFS(to, d + 1);
    if (flag) return;
  }
}
int main() {
  init();
  int n, u;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &is[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
  }
  for (int i = 1; i <= n; i++)
    if (!is[i] && a[i]) {
      if (cnt[i] == 1 && cnt[a[i]] == 1 && !is[a[i]]) newedge(i, a[i]);
    } else if (a[i] && cnt[a[i]] == 1)
      newedge(i, a[i]);
  for (int i = 1; i <= n; i++) cnt[i] = 0;
  ans = 0;
  int id;
  for (int i = 1; i <= n; i++)
    if (is[i]) {
      mx = 0;
      dep[i] = 1;
      dfs(i);
      if (mx > ans) id = i, ans = mx;
    }
  printf("%d\n", ans);
  path[1] = id;
  flag = 0;
  DFS(id, 2);
}

```