1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <vector>
 5 #include <cstring>
 6 #include <queue>
 7 using namespace std;
 8 const int maxn = 105;
 9 const int INF = 0x3f3f3f3f;
10 int n;
11 int energy[maxn];
12 int power[maxn];   //记录到每一点的力量值
13 int cnt[maxn];   //记录访问每一个点的次数
14 bool g[maxn][maxn];
15 bool reach[maxn][maxn];
16 void floyd()
17 {
18     for(int k = 1; k<=n; k++)
19         for(int i = 1; i<=n; i++)
20             for(int j = 1; j<=n; j++)
21             if(reach[i][j]||(reach[i][k]&&reach[k][j])) reach[i][j] = true;
22 }
23 bool spfa(int s)
24 {
25     queue<int> q;
26     memset(power,0,sizeof(power));
27     power[s] = 100;
28     memset(cnt,0,sizeof(power));
29     q.push(s);
30     while(!q.empty())
31     {
32         int v = q.front();
33         q.pop();
34         cnt[v]++;
35         if(cnt[v]>=n) return reach[v][n]; //有正环
36         for(int i = 1; i<=n; i++)
37         {
38             if(g[v][i]&&power[i]<power[v]+energy[i]&&power[v]+energy[i]>0)
39             {
40                 power[i] = power[v]+energy[i];
41 
42                 q.push(i);
43             }
44         }
45     }
46     return power[n]>0;
47 
48 }
49 void solve()
50 {
51     while(scanf("%d",&n)!=EOF&&n!=-1)
52     {
53         memset(g,false,sizeof(g));
54         memset(reach,false,sizeof(reach));
55         for(int i = 1; i<=n; i++)
56         {
57             int k,door;
58             scanf("%d %d",&energy[i],&k);
59             for(int j = 1; j<=k; j++)
60             {
61                 scanf("%d",&door);
62                 g[i][door] = true;
63                 reach[i][door] = true;
64             }
65         }
66         floyd();
67         if(!reach[1][n])
68         {
69             printf("hopeless\n");
70             continue;
71         }
72         if(spfa(1))
73         {
74             printf("winnable\n");
75         }
76         else
77         {
78             printf("hopeless\n");
79         }
80     }
81 }
82 int main()
83 {
84     solve();
85     return 0;
86 }