/**
          题意：    给出一个图，求最小顶点覆盖 
          做法：二分图最大匹配,最小顶点覆盖 == 最大匹配（双向图）/2；匈牙利算法，时间复杂度是O(VE),
**/
#include<iostream>
#include<cmath>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#define maxn 1570
using namespace std;
int g[maxn][maxn];
int linker[maxn];
int head[maxn];
int used[maxn];
int num = 0;
int n;
int tot = 0;
struct Node
{
          int to;
          int next;
}edge[maxn*10];
void Init()
{
          tot = 0;
          memset(head,-1,sizeof(head));
}
void addedge(int u,int v)
{
          edge[tot].to = v;
          edge[tot].next = head[u];
          head[u] = tot++;
}
bool dfs(int u)
{
          for(int v = head[u];~v;v = edge[v].next)
          {
                    int tt = edge[v].to;
                    if(used[tt] == 0 )
                    {
                              used[tt] = 1;
                              if(linker[tt] == -1||dfs(linker[tt]))
                              {
                                        linker[tt] = u;
                                        return true;
                              }
                    }
          }
          return false;
}
int hungary()
{
    memset(linker,-1,sizeof(linker));
    int res = 0;
    for(int i=0; i<n; i++)
    {
        memset(used,0,sizeof(used));
        if(dfs(i)) res++;
    }
    return res;
}
int main()
{
//#ifndef ONLINE_JUDGE
//    freopen("in.txt","r",stdin);
//#endif // ONLINE_JUDGE
    while(~scanf("%d",&n))
    {
        int u,v,Q;
        Init();
        memset(g,0,sizeof(g));
        for(int i=0; i<n; i++)
        {
                              scanf("%d:(%d)",&u,&Q);
                              while(Q--)
                              {
                                        scanf("%d",&v);
                                        addedge(u,v);
                                        addedge(v,u);
                              }
        }
        int res = hungary();
        printf("%d\n",res/2);
    }
}