#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
const int inf = 0xfffffff;
int outdeg[maxn];
vector<int> G[maxn];
int d[maxn];
int pre[maxn];
int q[3 * maxn];
bool vis[maxn];
int path[maxn];
int n;
void spfa() {
  int front = 0;
  int rear = 0;
  for (int i = 1; i <= n; ++i)
    if (d[i] == 1) q[rear++] = i;
  while (front < rear) {
    int x = q[front++];
    vis[x] = true;
    for (int i = 0; i < G[x].size(); ++i) {
      int y = G[x][i];
      if (!vis[y] && d[y] < d[x] + 1) {
        d[y] = d[x] + 1;
        q[rear++] = y;
        pre[y] = x;
      }
    }
  }
  int ret = 1;
  for (int i = 2; i <= n; ++i)
    if (d[ret] < d[i]) ret = i;
  printf("%d\n", d[ret]);
  int sz = d[ret];
  int tem = sz;
  int cur = ret;
  while (cur != -1) {
    path[--sz] = cur;
    cur = pre[cur];
  }
  printf("%d", path[tem - 1]);
  for (int i = tem - 2; i >= 0; --i) printf(" %d", path[i]);
  printf("\n");
}
int main() {
  while (scanf("%d", &n) == 1) {
    int tem;
    for (int i = 1; i <= n; ++i) {
      G[i].clear();
      scanf("%d", &tem);
      if (tem)
        d[i] = 1;
      else
        d[i] = -inf;
    }
    memset(outdeg, 0, sizeof(outdeg));
    for (int y = 1; y <= n; ++y) {
      int x;
      scanf("%d", &x);
      if (x == 0) continue;
      ++outdeg[x];
      G[y].push_back(x);
    }
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; ++i) {
      if (outdeg[i] > 1) {
        vis[i] = true;
        d[i] = -inf;
        G[i].clear();
      }
    }
    memset(pre, -1, sizeof(pre));
    spfa();
  }
}

### Prompt

Your challenge is to write 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;
const int maxn = 100000 + 10;
const int inf = 0xfffffff;
int outdeg[maxn];
vector<int> G[maxn];
int d[maxn];
int pre[maxn];
int q[3 * maxn];
bool vis[maxn];
int path[maxn];
int n;
void spfa() {
  int front = 0;
  int rear = 0;
  for (int i = 1; i <= n; ++i)
    if (d[i] == 1) q[rear++] = i;
  while (front < rear) {
    int x = q[front++];
    vis[x] = true;
    for (int i = 0; i < G[x].size(); ++i) {
      int y = G[x][i];
      if (!vis[y] && d[y] < d[x] + 1) {
        d[y] = d[x] + 1;
        q[rear++] = y;
        pre[y] = x;
      }
    }
  }
  int ret = 1;
  for (int i = 2; i <= n; ++i)
    if (d[ret] < d[i]) ret = i;
  printf("%d\n", d[ret]);
  int sz = d[ret];
  int tem = sz;
  int cur = ret;
  while (cur != -1) {
    path[--sz] = cur;
    cur = pre[cur];
  }
  printf("%d", path[tem - 1]);
  for (int i = tem - 2; i >= 0; --i) printf(" %d", path[i]);
  printf("\n");
}
int main() {
  while (scanf("%d", &n) == 1) {
    int tem;
    for (int i = 1; i <= n; ++i) {
      G[i].clear();
      scanf("%d", &tem);
      if (tem)
        d[i] = 1;
      else
        d[i] = -inf;
    }
    memset(outdeg, 0, sizeof(outdeg));
    for (int y = 1; y <= n; ++y) {
      int x;
      scanf("%d", &x);
      if (x == 0) continue;
      ++outdeg[x];
      G[y].push_back(x);
    }
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; ++i) {
      if (outdeg[i] > 1) {
        vis[i] = true;
        d[i] = -inf;
        G[i].clear();
      }
    }
    memset(pre, -1, sizeof(pre));
    spfa();
  }
}

```