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| // Renovations | |
| // Solution by Jacob Plachta | |
| using namespace std; | |
| template<typename T> T Abs(T x) { return(x<0 ? -x : x); } | |
| template<typename T> T Sqr(T x) { return(x*x); } | |
| string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); } | |
| const int INF = (int)1e9; | |
| const LD EPS = 1e-12; | |
| const LD PI = acos(-1.0); | |
| bool Read(int &x) | |
| { | |
| char c,r=0,n=0; | |
| x=0; | |
| for(;;) | |
| { | |
| c=GETCHAR(); | |
| if ((c<0) && (!r)) | |
| return(0); | |
| if ((c=='-') && (!r)) | |
| n=1; | |
| else | |
| if ((c>='0') && (c<='9')) | |
| x=x*10+c-'0',r=1; | |
| else | |
| if (r) | |
| break; | |
| } | |
| if (n) | |
| x=-x; | |
| return(1); | |
| } | |
| int N,K; | |
| int expMem[LIM]; | |
| int Exp(int a,int b) | |
| { | |
| if ((a<N) && (b==K) && (expMem[a])) | |
| return(expMem[a]); | |
| LL p=a,v=1; | |
| while (b) | |
| { | |
| if (b&1) | |
| v=v*p%MOD; | |
| p=p*p%MOD; | |
| b>>=1; | |
| } | |
| if ((a<N) && (b==K)) | |
| expMem[a]=v; | |
| return(v); | |
| } | |
| void Add(int &a,int b) | |
| { | |
| a=(a+b)%MOD; | |
| } | |
| int Prod(int a,int b) | |
| { | |
| return((LL)a*b%MOD); | |
| } | |
| void Div(int &a,int b) | |
| { | |
| a=Prod(a,Exp(b,MOD-2)); | |
| } | |
| // # of sequences with f forbidden nodes, r required nodes | |
| int Count(int f,int r) | |
| { | |
| int v=N-f-1; | |
| int c=Exp(v,K); | |
| if (!r) | |
| return(c); | |
| if (r==1) | |
| return((c-Exp(v-1,K)+MOD)%MOD); | |
| return((c-(LL)2*Exp(v-1,K)%MOD+Exp(v-2,K)+MOD)%MOD); | |
| } | |
| int main() | |
| { | |
| if (DEBUG) | |
| freopen("in.txt","r",stdin); | |
| // vars | |
| int T,t; | |
| int L,A,B; | |
| int i,dA,dB,dL; | |
| int P[LIM]={-1}; | |
| // testcase loop | |
| Read(T); | |
| Fox1(t,T) | |
| { | |
| // init | |
| Fill(expMem,0); | |
| // input | |
| Read(N),Read(K),Read(A),Read(B),A--,B--; | |
| Fox1(i,N-1) | |
| Read(P[i]),P[i]--; | |
| // compute depths of A, B, and their LCA | |
| queue<int> qA,qB; | |
| for (dA=0; A>=0; A=P[A],dA++) | |
| qA.push(A); | |
| for (dB=0; B>=0; B=P[B],dB++) | |
| qB.push(B); | |
| dL=max(dA,dB); | |
| for(;;) | |
| { | |
| if (qA.front()==qB.front()) | |
| break; | |
| dL--; | |
| if (Sz(qA)>dL) | |
| qA.pop(); | |
| if (Sz(qB)>dL) | |
| qB.pop(); | |
| } | |
| dL--,dA--,dB--; | |
| // consider all possible final states | |
| int ans=0; | |
| if (!dL) // original LCA is the root? | |
| { | |
| Fox(A,dA+1) | |
| Fox(B,dB+1) | |
| { | |
| if ((dA) && (!A) || (dB) && (!B)) | |
| continue; | |
| Add(ans,Prod(A+B,Count(A+B-(A?1:0)-(B?1:0),(A<dA?1:0)+(B<dB?1:0)))); | |
| } | |
| } | |
| else | |
| { | |
| // consider LCA being retained | |
| Fox1(L,dL) | |
| Add(ans,Prod(dA+dB-2*dL,Count(dA+dB-2*dL+L-1,(L<dL ? 1 : 0)))); | |
| // consider LCA being broken | |
| Fox1(A,dA) | |
| Fox1(B,dB) | |
| { | |
| if ((A>dA-dL) && (B>dB-dL)) | |
| continue; | |
| Add(ans,Prod(A+B,Count(A+B-2,(A<dA?1:0)+(B<dB?1:0)))); | |
| } | |
| } | |
| // divide by total number of sequences | |
| Div(ans,Exp(N-1,K)); | |
| // output | |
| printf("Case #%d: %d\n",t,ans); | |
| } | |
| return(0); | |
| } |