Problem: Isaac H. Ives attended an international student party and made a lot of girl friends (as many other persons expected). To strike up a good friendship with them, he decided to have dates with them. However, it is hard for him to schedule dates because he made so many friends. Thus he decided to find the best schedule using a computer program. The most important criterion in scheduling is how many different girl friends he will date. Of course, the more friends he will date, the better the schedule is. However, though he has the ability to write a program finding the best schedule, he doesn’t have enough time to write it.

Your task is to write a program to find the best schedule instead of him.



Input

The input consists of a series of data sets. The first line of each data set contains a single positive integer N (N ≤ 1,000) that represents the number of persons Isaac made friends with in the party. The next line gives Isaac’s schedule, followed by N lines that give his new friends’ schedules. Each schedule consists of a positive integer M that represents the number of available days followed by M positive integers each of which represents an available day.

The input is terminated by a line that contains a single zero. This is not part of data sets and should not be processed.

Output

For each data set, print a line that contains the maximum number of girl friends Isaac can have dates with.

Example

Input

3
3 1 3 5
2 1 4
4 1 2 3 6
1 3
0


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <vector>
#include <list>
#include <cmath>
#include <fstream>
#include <algorithm>
#include <string>
#include <queue>
#include <set>
#include <map>
#include <complex>
#include <iterator>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <stack>
#include <climits>
#include <deque>
#include <bitset>
#include <cassert>
#include <ctime>
using namespace std;
 
typedef long long ll;
typedef pair<int,int> pii;
const int dy[]={-1,0,1,0},dx[]={0,1,0,-1};
// adjust problem by problem
const double EPS=1e-8;
const double PI=acos(-1.0);
#ifdef __GNUC__
int popcount(int n){return __builtin_popcount(n);}
int popcount(ll n){return __builtin_popcountll(n);}
#endif
#ifndef __GNUC__
template<class T> int popcount(T n){int cnt=0;while(n){if(n%2)cnt++;n/=2;}return cnt;}
#endif
template<class T>int SIZE(T a){return a.size();}
template<class T>string IntToString(T num){string res;stringstream ss;ss<<num;return ss.str();}
template<class T>T StringToInt(string str){T res=0;for(int i=0;i<SIZE(str);i++)res=(res*10+str[i]-'0');return res;}
template<class T>T gcd(T a,T b){if(b==0)return a;return gcd(b,a%b);}
template<class T>T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> void PrintSeq(T &a,int sz){for(int i=0;i<sz;i++){cout<<a[i];if(sz==i+1)cout<<endl;else cout<<' ';}}
bool EQ(double a,double b){return abs(a-b)<EPS;}
void fastStream(){cin.tie(0);std::ios_base::sync_with_stdio(0);}
vector<string> split(string str,char del){
  vector<string> res;
  for(int i=0,s=0;i<SIZE(str);i++){
    if(str[i]==del){if(i-s!=0)res.push_back(str.substr(s,i-s));s=i+1;}
    else if(i==SIZE(str)-1){res.push_back(str.substr(s));}
  }
  return res;
}
 
class Dinic{
private:
  static const int MAX_V=200001;
  static const ll INF=1000000000000000LL;
  class edge{
  public:
    int to,rev;
    ll cap;
    edge(int to_,ll cap_,int rev_){
      to=to_;
      cap=cap_;
      rev=rev_;
    }
  };
public:
  vector<edge> G[MAX_V];
  // sからの距離
  int level[MAX_V];
  // どこまで調べ終わったか
  int iter[MAX_V];
  void add_edge(int from,int to,ll cap){
    G[from].push_back(edge(to,cap,G[to].size()));
    G[to].push_back(edge(from,0,G[from].size()-1));
  }
  void init(){
    for(int i=0;i<MAX_V;i++)G[i].clear();
  }
private:
  // sからの最短距離を求める
  void bfs(int s){
    memset(level,-1,sizeof(level));
    queue<int> que;
    level[s]=0;
    que.push(s);
    while(!que.empty()){
      int v=que.front();que.pop();
      for(int i=0;i<(int)G[v].size();i++){
        edge &e=G[v][i];
        if(e.cap>0&&level[e.to]<0){
          level[e.to]=level[v]+1;
          que.push(e.to);
        }
      }
    }
  }
  // 増加パスをdfsで探す
  ll dfs(int v,int t,ll f){
    if(v==t)return f;
    for(int &i=iter[v];i<(int)G[v].size();i++){
      edge &e=G[v][i];
      if(e.cap>0&&level[v]<level[e.to]){
        ll d=dfs(e.to,t,min(f,e.cap));
        if(d>0){
          e.cap-=d;
          G[e.to][e.rev].cap+=d;
          return d;
        }
      }
    }
    return 0;
  }
public:
  // sからtへの最大流を求める
  ll max_flow(int s,int t){
    ll flow=0;
    while(1){
      bfs(s);
      if(level[t]<0)return flow;
      memset(iter,0,sizeof(iter));
      ll f;
      while((f=dfs(s,t,INF))>0)flow+=f;
    }
  }
};
 
Dinic dinic;
int main(){
  int N;
  while(cin>>N&&N){
    dinic.init();
    vector<ll> ss;
    map<ll,int> dayIdx;
    int daySize=0;
    vector<ll> fs[1001];
    {
      int M;
      cin>>M;
      for(int i=0;i<M;i++){
        ll a;
        cin>>a;
        ss.push_back(a);
      }
      sort(ss.begin(),ss.end());
      ss.erase(unique(ss.begin(),ss.end()),ss.end());
      for(int i=0;i<(int)ss.size();i++){
        ll a=ss[i];
        if(dayIdx.count(a)==0)dayIdx[a]=daySize++;
        ss[i]=dayIdx[a];
      }
    }
    for(int i=0;i<N;i++){
      int M;
      cin>>M;
      for(int j=0;j<M;j++){
        ll a;
        cin>>a;
        fs[i].push_back(a);
        // if(dayIdx.count(a)!=0)fs[i][j]=dayIdx[a];
        // else
        //   fs[i][j]=-1;
      }
      sort(fs[i].begin(),fs[i].end());
      fs[i].erase(unique(fs[i].begin(),fs[i].end()),fs[i].end());
      for(int j=0;j<(int)fs[i].size();j++){
        ll a=fs[i][j];
        if(dayIdx.count(a)!=0)fs[i][j]=dayIdx[a];
        else
          fs[i][j]=-1;
      }
    }
    int node=(int)dayIdx.size();
    const int st=node++;
    const int ed=node++;
    for(int j=0;j<(int)ss.size();j++){
      int a=ss[j];
      dinic.add_edge(st,a,1);
    }
    for(int i=0;i<N;i++){
      const int to=i+node;
      for(int j=0;j<(int)fs[i].size();j++){
        const int from=fs[i][j];
        if(from!=-1)
          dinic.add_edge(from,to,1);
      }
      dinic.add_edge(to,ed,1);
    }
    node+=N;
    printf("%d\n",(int)dinic.max_flow(st,ed));
    //cout<<dinic.max_flow(st,ed)<<endl;
  }
  return 0;
}