1 /* */
  2 # include <iostream>
  3 # include <stdio.h>
  4 # include <string.h>
  5 # include <algorithm>
  6 # include <bitset>
  7 # include <ctime>
  8 # include <cmath>
  9 # include <list>
 10 # include <cctype>
 11 # include <cassert>
 12 # include <iomanip>
 13 # include <deque>
 14 # include <queue>
 15 # include <stack>
 16 # include <set>
 17 # include <map>
 18 # include <vector>
 19 using namespace std;
 20
 21 const int inf=0x3f3f3f3f;
 22 const int maxn=2e5+50;
 23 int a[maxn];
 24 int zt[maxn];
 25 vector<int>ve[maxn];
 26
 27 void dfs(int x, int y)///x此结点,y前驱结点的权值
 28 {
 29     if( zt[x]==-2 )///有两个前驱,(既然已经判断出有两个前驱,且此路之前来过,就没必要再往下遍历了)
 30         return ;
 31
 32
 33     if( !zt[x] )///还没有访问过
 34         zt[x] = y;///记录其前驱节点的权值
 35     else///已经访问过之后又来访问说明有两个前驱结点了,标记一下
 36         zt[x] = -2;
 37
 38
 39     if( a[x] )///此结点的权值不为0
 40         y=a[x];
 41
 42
 43     for( auto it:ve[x] )
 44     {
 45         if( zt[it]!=y )///之前的前驱结点的值不为现在前驱结点的值,
 46             ///若zt[it]原来为0,说明没有遍历过
 47             ///若zt[it]原来为-1,或者是一个正整数
 48             ///但之前前驱结点的值与此时前驱节点的值不一样
 49             ///那么就再进行一次遍历
 50             ///至于具体是什么情况,for有前面的判断决定
 51         {
 52             dfs(it, y);
 53         }
 54     }
 55
 56 }
 57
 58 int main()
 59 {
 60     memset(zt, 0, sizeof(zt));
 61     int n;
 62     cin>>n;
 63     for(int i=1; i<=n; i++ )
 64     {
 65         cin>>a[i];
 66         int x;
 67         cin>>x;
 68         for(int j=1; j<=x; j++ )
 69         {
 70             int y;
 71             cin>>y;
 72             ve[i].push_back(y);
 73         }
 74     }
 75     dfs(1, -1);
 76
 77     for(int i=1; i<=n; i++ )
 78     {
 79         if( zt[i]==-1 )
 80             cout<<"sober ";
 81
 82         else if( zt[i]==-2 )
 83             cout<<"unknown ";
 84
 85         else
 86             cout<<zt[i]<<' ';
 87
 88         if( a[i]>0 )
 89             cout<<a[i]<<endl;
 90
 91         else
 92         {
 93             if( zt[i]==-1 )
 94                 cout<<"sober"<<endl;
 95
 96             else if( zt[i]==-2 )
 97                 cout<<"unknown"<<endl;
 98
 99             else
100                 cout<<zt[i]<<endl;
101         }
102     }
103     return 0;
104 }