#include <bits/stdc++.h>
const int N = 100005;
const int inf = 1 << 28;
const double eps = 1e-8;
using namespace std;
int n;
int a[N];
int tot, pre[N];
vector<int> g[N];
void init() {
  for (int i = 1; i <= n; i++) g[i].clear();
}
void input() {
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int x;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x) g[x].push_back(i);
  }
}
int res, dp[N];
void dfs(int u) {
  dp[u] = 0;
  if (a[u]) {
    dp[u] = 1;
    return;
  }
  int sz = g[u].size();
  if (sz > 1) {
    dp[u] = -inf;
    return;
  }
  for (int i = 0; i < sz; i++) {
    int v = g[u][i];
    if (dp[v] == -1) dfs(v);
    dp[u] = max(dp[u], dp[v]);
  }
  if (!dp[u]) dp[u] = -inf;
  dp[u]++;
}
void find(int u) {
  int sz = g[u].size();
  printf("%d", u);
  if (a[u])
    puts("");
  else
    putchar(' ');
  if (a[u]) return;
  for (int i = 0; i < sz; i++) {
    int v = g[u][i];
    if (dp[u] == dp[v] + 1) {
      find(v);
      return;
    }
  }
}
void work() {
  memset(dp, -1, sizeof(dp));
  res = 0;
  int k = -1;
  for (int i = 1; i <= n; i++) {
    if (dp[i] == -1) dfs(i);
    if (res < dp[i]) res = dp[i], k = i;
  }
  printf("%d\n", res);
  find(k);
}
int main() {
  while (scanf("%d", &n) > 0) {
    init();
    input();
    work();
  }
  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>
const int N = 100005;
const int inf = 1 << 28;
const double eps = 1e-8;
using namespace std;
int n;
int a[N];
int tot, pre[N];
vector<int> g[N];
void init() {
  for (int i = 1; i <= n; i++) g[i].clear();
}
void input() {
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int x;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x) g[x].push_back(i);
  }
}
int res, dp[N];
void dfs(int u) {
  dp[u] = 0;
  if (a[u]) {
    dp[u] = 1;
    return;
  }
  int sz = g[u].size();
  if (sz > 1) {
    dp[u] = -inf;
    return;
  }
  for (int i = 0; i < sz; i++) {
    int v = g[u][i];
    if (dp[v] == -1) dfs(v);
    dp[u] = max(dp[u], dp[v]);
  }
  if (!dp[u]) dp[u] = -inf;
  dp[u]++;
}
void find(int u) {
  int sz = g[u].size();
  printf("%d", u);
  if (a[u])
    puts("");
  else
    putchar(' ');
  if (a[u]) return;
  for (int i = 0; i < sz; i++) {
    int v = g[u][i];
    if (dp[u] == dp[v] + 1) {
      find(v);
      return;
    }
  }
}
void work() {
  memset(dp, -1, sizeof(dp));
  res = 0;
  int k = -1;
  for (int i = 1; i <= n; i++) {
    if (dp[i] == -1) dfs(i);
    if (res < dp[i]) res = dp[i], k = i;
  }
  printf("%d\n", res);
  find(k);
}
int main() {
  while (scanf("%d", &n) > 0) {
    init();
    input();
    work();
  }
  return 0;
}

```