#include <bits/stdc++.h>
using namespace std;
int n;
const int maxn = 1e5 + 10;
vector<int> G[maxn];
int c[maxn], out[maxn], d[maxn], fa[maxn];
int t, ss, tt, mlen;
void bfs(int x) {
  queue<int> q;
  q.push(x);
  memset(d, -1, sizeof(d));
  d[x] = mlen = 0;
  fa[x] = -1;
  t = x;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (d[v] != -1 || c[v] == -1) continue;
      if (out[v] != 1) continue;
      d[v] = d[u] + 1;
      fa[v] = u;
      q.push(v);
      if (d[v] > mlen) {
        mlen = d[v];
        t = v;
      }
    }
  }
}
int main() {
  memset(out, 0, sizeof(out));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &t);
    if (t) {
      G[i].push_back(t);
      G[t].push_back(i);
      out[t]++;
    }
  }
  int ans = -1;
  for (int i = 1; i <= n; i++) {
    if (c[i]) {
      bfs(i);
      if (d[t] > ans) {
        ans = d[t];
        ss = i;
        tt = t;
      }
    }
  }
  bfs(ss);
  printf("%d\n", ans + 1);
  while (tt != -1) {
    printf("%d ", tt);
    tt = fa[tt];
  }
  return 0;
}

### 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>
using namespace std;
int n;
const int maxn = 1e5 + 10;
vector<int> G[maxn];
int c[maxn], out[maxn], d[maxn], fa[maxn];
int t, ss, tt, mlen;
void bfs(int x) {
  queue<int> q;
  q.push(x);
  memset(d, -1, sizeof(d));
  d[x] = mlen = 0;
  fa[x] = -1;
  t = x;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (d[v] != -1 || c[v] == -1) continue;
      if (out[v] != 1) continue;
      d[v] = d[u] + 1;
      fa[v] = u;
      q.push(v);
      if (d[v] > mlen) {
        mlen = d[v];
        t = v;
      }
    }
  }
}
int main() {
  memset(out, 0, sizeof(out));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &t);
    if (t) {
      G[i].push_back(t);
      G[t].push_back(i);
      out[t]++;
    }
  }
  int ans = -1;
  for (int i = 1; i <= n; i++) {
    if (c[i]) {
      bfs(i);
      if (d[t] > ans) {
        ans = d[t];
        ss = i;
        tt = t;
      }
    }
  }
  bfs(ss);
  printf("%d\n", ans + 1);
  while (tt != -1) {
    printf("%d ", tt);
    tt = fa[tt];
  }
  return 0;
}

```