Codeforces Round #203 (Div. 2)B Resort

Resort

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 350B

Description

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 nobjects (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, ..., v_k (k ≥ 1) and meet the following conditions:

1. Objects with numbers v₁, v₂, ..., v_k - 1 are mountains and the object with number v_k is the hotel.
2. For any integer i(1 ≤ i < k), there is exactly one ski track leading from object v_i. This track goes to object v_i + 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, ..., type_n — the types of the objects. If type_i equals zero, then the i-th object is the mountain. If type_i 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, ..., a_n (0 ≤ a_i ≤ n) — the description of the ski tracks. If number a_i equals zero, then there is no such object v, that has a ski track built from v to i. If number a_i doesn't equal zero, that means that there is a track built from object a_i 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, ..., v_k — the path. If there are multiple solutions, you can print any of them.

Sample Input

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

/*
竟然看漏了一个隐藏条件，每个点的入度只能为零
*/
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int val[100005];
vector<int> edge[100005];
int vis[100005];
vector<int> path[100005];
int n;
int a;
//每个点只可能有一个入度
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&val[i]);
    }
    for(int i=1;i<=n;i++){
        scanf("%d",&a);
        if(a){
            edge[i].push_back(a);//建图
            vis[a]++;
        }
    }
    //从1开始向前找
    for(int i=1;i<=n;i++){
        if(val[i]){
            path[i].push_back(i);
            if(edge[i].size()==0){
                continue;
            }
            int pos=edge[i][0];
            while(true){
                if(val[pos]==1){//如果这一步是1的话肯定是不行的
                    break;
                }
                if(vis[pos]>1){//如果这一步有两个出度也是不行的
                    break;
                }
                if(edge[pos].size()==0){//如果没有下一步了
                    path[i].push_back(pos);
                    break;
                }
                path[i].push_back(pos);
                pos=edge[pos][0];
            }
        }
    }
    int maxn=-1;
    for(int i=1;i<=n;i++){
        maxn=max((int)path[i].size(),maxn);
    }
    for(int i=1;i<=n;i++){
        if(path[i].size()==maxn){
            printf("%d
",maxn);
            for(int j=path[i].size()-1;j>=0;j--){
                printf(j==path[i].size()-1?"%d":" %d",path[i][j]);
            }
            printf("
");
            break;
        }
    }
    return 0;
}