#include <bits/stdc++.h>
const int N = 100001;
int sz[N], tp[N], v[N], pre[N];
int f[N], ml[N], cnt;
bool mem[N];
struct Edge {
  int to, next, w;
} eg[N << 1];
void add(int x, int y) {
  eg[cnt].to = y;
  eg[cnt].next = v[x];
  v[x] = cnt++;
}
void dfs(int x) {
  int i;
  mem[x] = 1;
  if (tp[x] == 1 && (ml[x] = 1)) return;
  if (sz[x] != 1) return;
  for (i = v[x]; i != -1; i = eg[i].next) {
    if (!mem[eg[i].to]) dfs(eg[i].to);
    if (ml[eg[i].to] && ml[eg[i].to] + 1 > ml[x])
      ml[x] = ml[eg[i].to] + 1, f[x] = eg[i].to;
  }
}
int main() {
  int n, m, i, j, k, x, y, t, rl, rx;
  while (~scanf("%d", &n)) {
    memset(sz, 0, sizeof(sz));
    memset(v, -1, sizeof(v));
    memset(mem, 0, sizeof(mem));
    memset(ml, 0, sizeof(ml));
    memset(f, 0, sizeof(f));
    for (i = 1; i <= n; i++) scanf("%d", &tp[i]);
    for (cnt = 0, i = 1; i <= n; i++) {
      scanf("%d", &x);
      sz[x]++;
      add(x, i);
    }
    for (rl = rx = 0, i = 1; i <= n; i++) {
      dfs(i);
      if (ml[i] > rl) rx = i, rl = ml[i];
    }
    printf("%d\n", rl);
    for (i = rx; i != 0; i = f[i]) printf("%d ", i);
    printf("\n");
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 N = 100001;
int sz[N], tp[N], v[N], pre[N];
int f[N], ml[N], cnt;
bool mem[N];
struct Edge {
  int to, next, w;
} eg[N << 1];
void add(int x, int y) {
  eg[cnt].to = y;
  eg[cnt].next = v[x];
  v[x] = cnt++;
}
void dfs(int x) {
  int i;
  mem[x] = 1;
  if (tp[x] == 1 && (ml[x] = 1)) return;
  if (sz[x] != 1) return;
  for (i = v[x]; i != -1; i = eg[i].next) {
    if (!mem[eg[i].to]) dfs(eg[i].to);
    if (ml[eg[i].to] && ml[eg[i].to] + 1 > ml[x])
      ml[x] = ml[eg[i].to] + 1, f[x] = eg[i].to;
  }
}
int main() {
  int n, m, i, j, k, x, y, t, rl, rx;
  while (~scanf("%d", &n)) {
    memset(sz, 0, sizeof(sz));
    memset(v, -1, sizeof(v));
    memset(mem, 0, sizeof(mem));
    memset(ml, 0, sizeof(ml));
    memset(f, 0, sizeof(f));
    for (i = 1; i <= n; i++) scanf("%d", &tp[i]);
    for (cnt = 0, i = 1; i <= n; i++) {
      scanf("%d", &x);
      sz[x]++;
      add(x, i);
    }
    for (rl = rx = 0, i = 1; i <= n; i++) {
      dfs(i);
      if (ml[i] > rl) rx = i, rl = ml[i];
    }
    printf("%d\n", rl);
    for (i = rx; i != 0; i = f[i]) printf("%d ", i);
    printf("\n");
  }
  return 0;
}

```