// 532ms
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
#include<stack>
#include<set>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define pb push_back
template<typename T>
void read(T &m)
{T x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}m = x*f;
}
template<typename T>
void out(T a)
{if(a>9) out(a/10);putchar(a%10+'0');
}
const int N = 2020<<1;//倍增点数
const int M = 202200;
int head[M],tot;
struct edge{int to,w,Next;
}e[M];
void ins(int a,int b,int w = 0)
{e[++tot].Next = head[a];e[tot].to = b;e[tot].w = w;head[a] = tot;
}
int pre[N],dfs_clock,low[N];
int belong[N],scc,n;
stack<int> S;
bool stk[N];
void Tarjan(int u)
{pre[u] = low[u] = ++dfs_clock;S.push(u);stk[u] = true;int v;//点u所在连通分量的出度；for(int i = head[u];i;i = e[i].Next){v = e[i].to;if(pre[v] == 0){Tarjan(v);low[u] = min(low[u],low[v]);}else if(stk[v]){low[u] = min(low[u],pre[v]);}}if(pre[u] == low[u]){//强连通分量的根节点++scc;do{v = S.top();S.pop();stk[v] = false;//if(v <= n)belong[v] = scc;//else vec[scc].pb(v);}while(v != u);}
}
int ans[N];
int main()
{int v,T,kase = 1;read(n);rep1(u,1,n){int k;read(k);rep0(j,0,k){read(v);ins(u,v+n);//妹子标号要加上n;
        }}rep1(u,1,n){read(v);ins(v+n,u);//反向边***
    }rep1(u,1,n)if(pre[u] == 0)Tarjan(u);rep1(u,1,n){int cnt = 0;for(int i = head[u];i;i = e[i].Next){//遍历边v=e[i].to;if(belong[u] == belong[v])   //同一个强连通分量ans[cnt++] = v-n;}sort(ans,ans+cnt);out(cnt);rep0(i,0,cnt){putchar(' ');out(ans[i]);}puts("");}return 0;
}