/*

题意：给出一个有向强连通图，每条边有两个值分别是破坏该边的代价和把该边建成无向边的代价（建立无向边的前提是删除该边）问是否存在一个集合S，和一个集合的补集T，破坏所有S集合到T集合的边代价和是X，然后修复T到S的边为无向边代价和是Y，满足Y
  
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        while(t--) {
                init();
            scanf("%d%d",&n,&m);
        S=0;T=n+1;
        memset(w,0,sizeof(w));
            for(i=1;i<=m;i++) {
                scanf("%d%d%d%d",&a,&b,&c,&d);
                add(a,b,d);
               w[b]+=c;
               w[a]-=c;
            }
            sum=0;
            for(i=1;i<=n;i++) {
                if(w[i]>0) {
                    sum+=w[i];
                    add(S,i,w[i]);
                }
                if(w[i]<0)
                    add(i,T,-w[i]);
            }
            ans=dinic(S,T);
            if(sum==ans)
                printf("Case #%d: happy\n",++k);
            else
                printf("Case #%d: unhappy\n",++k);
        }

return 0;
}