#include using namespace std; using ll=long long; const int N=255,P=998244353; inline ll ksm(ll a,int b){ ll res=1; while(b){ if(b&1)res=res*a%P; a=a*a%P,b>>=1; } return res; } inline ll ginv(int x){return ksm(x,P-2);} int n; int d[N][3]; int kp,kq; int a[N][N]; ll val[N]; inline int mol(int x,int y){return x+y>=P?x+y-P:x+y;} int x,y; struct Line{ int v[N];//v[i]->f[0][i] inline int& operator [](int x){return v[x];} inline const int& operator [](int x)const{return v[x];} Line(){memset(v,0,sizeof v);} inline Line operator +(const Line B){ Line C; for(int i=0;i<=y+1;i++)C[i]=mol(v[i],B[i]); return C; } inline void operator +=(const Line B){*this=*this+B;} inline Line operator -(const Line B){ Line C; for(int i=0;i<=y+1;i++)C[i]=mol(v[i],P-B[i]); return C; } inline void operator -=(const Line B){*this=*this-B;} inline Line operator *(int k){ Line C; for(int i=0;i<=y+1;i++)C[i]=1ll*v[i]*k%P; return C; } inline void operator *=(int k){(*this)=(*this)*k;} inline Line operator /(int k){ k=ginv(k); if(k==0)assert(0); return (*this)*k; } inline void operator /=(int k){(*this)=(*this)/k;} inline ll gt(){ ll res=v[y+1]; for(int i=0;i<=y;i++)res+=val[i]*v[i]%P; return res%P; } }rea[N][N]; int main(){ ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); cin>>n;kp=n*(n-1)/2-(n-2); int u,v,w; for(int i=1;i>u>>v>>w; x+=(w==1),y+=(w==2); d[u][w]++,d[v][w]++; } for(int i=0;i<=y;i++)rea[0][i].v[i]=1; ll tp=(P-ginv(n)*kp%P*(x+2*y)%P)%P; for(int i=0;i<=x;i++){ for(int j=0;j<=y;j++){ if(x==i&&y==j)continue; kq=n-i-j; Line cur=rea[i][j]*((1ll*kp*(x+2*y)-1ll*kq*(i+2*j)-1ll*(kp-kq+1)*((x-i)+2*(y-j)))%P+P); if(i)cur-=rea[i-1][j]*(1ll*(kp-kq)*i%P); if(j)cur-=rea[i][j-1]*(2ll*(kp-kq)*j%P); if(ja[r][i])r=j; } swap(a[i],a[r]); ll inv=ginv(a[i][i]); for(int j=i;j<=y+1;j++)a[i][j]=a[i][j]*inv%P; for(int j=0;j<=y;j++){ if(i==j)continue; inv=a[j][i]; for(int k=i;k<=y+1;k++)a[j][k]=mol(a[j][k],P-inv*a[i][k]%P); } } for(int i=0;i<=y;i++)val[i]=a[i][y+1]; ll ans=0; for(int i=1;i<=n;i++){ if(d[i][1]==x&&d[i][2]==y){ cout<<"0\n"; return 0; } ans+=rea[d[i][1]][d[i][2]].gt(); } cout<<(ans%P+P)%P<<"\n"; return 0; }