CF350B Resort（dfs）

原题链接：https://codeforces.com/problemset/problem/350/B

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, ..., 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_kis 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.

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
思路：反向建边，因为对于每一个点能够到达这个点的只有一个，类似于一棵树，从反向跑能保证路径唯一，然后就记录从每一个酒店跑的最远距离，还有一些细节，不能同时经过两个酒店

#include<bits/stdc++.h>
#define N 100005
using namespace std;
int n,x,ans=-1;
int b[N],a[N],vis[N];
vector<int>e[N];
vector<int>dis[N];
//每一个点只可能有一条路劲到达他，但他可能可以通过多条路径到达别的点
int main()
{
	memset(vis,0,sizeof(vis));
	memset(b,0,sizeof(b));
	scanf("%d",&n);
	for(int i = 1; i <= n; i++) scanf("%d",&b[i]);
	for(int i = 1; i <= n; i++) {
		scanf("%d",&x);
		if(x != 0) {
			e[i].push_back(x);//反向建图，从酒店开始走
			vis[x]++;
		}
	}
	for(int i=1;i<=n;i++) {
		if(b[i]!=0) {
			dis[i].push_back(i);
			if(e[i].size() == 0) continue;
			int y = e[i][0];
			while(1) {
				if(b[y] == 1)  break;  //表示下一步也是酒店，那就不行
				if(vis[y] > 1) break; //表示有多条路径到达y点，那么反过来就是从y点可以走到酒店，也可以走别的路
				if(e[y].size() == 0)   //表示没有下一步了
				{
					dis[i].push_back(y);
					break;
				}
				dis[i].push_back(y);
				y = e[y][0];
			}
		}
	}
	for(int i=1;i<=n;i++)
		ans=max(ans,(int)dis[i].size());
	printf("%d\n",ans);
	for(int i=1;i<=n;i++)
	{
		if((int)dis[i].size() == ans)
		{
			for(int j=dis[i].size()-1;j >= 0; j--)
				printf("%d ",dis[i][j]);
			break;
		}
	}
	return 0;
}