1 //2017-08-20
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <iostream>
  5 #include <algorithm>
  6 #include <cmath>
  7 #include <vector>
  8 
  9 using namespace std;
 10 
 11 const int N = 110;
 12 int n, in_degree[N], out_degree[N];
 13 bool vis[N];
 14 vector<int> G[N];
 15 vector<int> rG[N];
 16 vector<int> vs;
 17 int cmp[N];
 18 
 19 void add_edge(int u, int v){
 20     G[u].push_back(v);
 21     rG[v].push_back(u);
 22 }
 23 
 24 //input: u 顶点
 25 //output: vs 后序遍历顺序的顶点列表
 26 void dfs(int u){
 27     vis[u] = true;
 28     for(int i = 0; i < G[u].size(); i++){
 29         int v = G[u][i];
 30         if(!vis[v])
 31           dfs(v);
 32     }
 33     vs.push_back(u);
 34 
 35 }
 36 
 37 //input: u 顶点编号; k 拓扑序号
 38 //output: cmp[] 强连通分量拓扑序
 39 void rdfs(int u, int k){
 40     vis[u] = true;
 41     cmp[u] = k;
 42     for(int i = 0; i < rG[u].size(); i++){
 43         int v = rG[u][i];
 44         if(!vis[v])
 45           rdfs(v, k);
 46 
 47     }
 48 
 49 }
 50 
 51 //Strongly Connected Component 强连通分量
 52 //input: n 顶点个数
 53 //output: k 强连通分量数;
 54 int scc(){
 55     memset(vis, 0, sizeof(vis));
 56     vs.clear();
 57     for(int u = 1; u <= n; u++)
 58       if(!vis[u]){
 59           dfs(u);
 60       }
 61     int k = 0;
 62     memset(vis, 0, sizeof(vis));
 63     for(int i = vs.size()-1; i >= 0; i--)
 64       if(!vis[vs[i]])
 65         rdfs(vs[i], k++);
 66     return k;
 67 }
 68 
 69 void solve(){
 70     int k = scc();
 71     int ans = 0;
 72     memset(in_degree, 0, sizeof(in_degree));
 73     memset(out_degree, 0, sizeof(out_degree));
 74     for(int u = 1; u <= n; u++){
 75         memset(vis, 0, sizeof(vis));
 76         for(int i = 0; i < G[u].size(); i++){
 77             int v = G[u][i];
 78             if(vis[v])continue;
 79             vis[v] = 1;
 80             if(cmp[u] != cmp[v]){
 81                 out_degree[cmp[u]]++;
 82                 in_degree[cmp[v]]++;
 83             }
 84         }
 85     }
 86     int a = 0, b = 0;
 87     for(int i = 0; i < k; i++){
 88           if(in_degree[i] == 0)a++;
 89         if(out_degree[i] == 0)b++;
 90     }
 91     ans = max(a, b);
 92     if(k == 1)ans = 0;
 93     printf("%d\n%d\n", a, ans);
 94 }
 95 
 96 int main()
 97 {
 98     while(scanf("%d", &n)!=EOF){
 99         for(int u = 1; u <= n; u++){
100             G[u].clear();
101             rG[u].clear();
102         }
103         int v;
104         for(int u =1; u <= n; u++){
105             while(scanf("%d", &v)==1 && v){
106                 add_edge(u, v);
107             }
108         }
109         solve();
110     }
111 
112     return 0;
113 }