// Log Drivin' Hirin' // Solution by Jacob Plachta #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define LL long long #define LD long double #define PR pair #define Fox(i,n) for (i=0; i=0; i--) #define FoxR1(i,n) for (i=n; i>0; i--) #define FoxRI(i,a,b) for (i=b; i>=a; i--) #define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++) #define Min(a,b) a=min(a,b) #define Max(a,b) a=max(a,b) #define Sz(s) int((s).size()) #define All(s) (s).begin(),(s).end() #define Fill(s,v) memset(s,v,sizeof(s)) #define pb push_back #define mp make_pair #define x first #define y second template T Abs(T x) { return(x < 0 ? -x : x); } template 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); #define GETCHAR getchar_unlocked 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); } struct Line { mutable LL k, m, p; bool operator<(const Line& o) const { return k < o.k; } bool operator<(LL x) const { return p < x; } }; struct LineContainer : multiset> { const LL inf = LLONG_MAX; LL div(LL a, LL b) { // floored division return a / b - ((a ^ b) < 0 && a % b); } bool isect(iterator x, iterator y) { if (y == end()) { x->p = inf; return false; } if (x->k == y->k) x->p = x->m > y->m ? inf : -inf; else x->p = div(y->m - x->m, x->k - y->k); return x->p >= y->p; } void add(LL k, LL m) { auto z = insert({ k, m, 0 }), y = z++, x = y; while (isect(y, z)) z = erase(z); if (x != begin() && isect(--x, y)) isect(x, y = erase(y)); while ((y = x) != begin() && (--x)->p >= y->p) isect(x, erase(y)); } LL query(LL x) { assert(!empty()); auto l = *lower_bound(x); return l.k * x + l.m; } }; #define LIM 1000001 #define MOD 1000000007 int N, M, K, ans; int P[LIM], L[LIM], H[LIM], X[LIM], Y[LIM]; vector ch[LIM], Q[LIM]; int nid, id[LIM]; LineContainer LC[LIM]; void ReadSeq(int* V, int N, int K, int mod) { int i, A, B, C; Fox(i, K) Read(V[i]); Read(A), Read(B), Read(C); if (!mod) Read(mod); FoxI(i, K, N - 1) V[i] = ((LL)A * V[i - 2] + (LL)B * V[i - 1] + C) % (mod < 0 ? i : mod) + 1; } LL GetQueryAns(int i, LL d, int c) { return(c * d + LC[id[i]].query(c)); } void rec(int i, LL d) { // init new line container id[i] = nid++; // recurse to children for (auto c : ch[i]) { rec(c, d + L[c]); // merge smaller line container into larger one int& a = id[i], & b = id[c]; if (Sz(LC[a]) < Sz(LC[b])) swap(a, b); for (auto ln : LC[b]) LC[a].add(ln.k, ln.m); LC[b].clear(); } // add line for hireable log driver or leaf node LC[id[i]].add(-d, Sz(ch[i]) ? GetQueryAns(i, d, H[i]) : 0); // answer queries at this node for (auto c : Q[i]) ans = ans * ((-GetQueryAns(i, d, c) + 1) % MOD) % MOD; ch[i].clear(), Q[i].clear(); } int ProcessCase() { int i; // input Read(N), Read(M), Read(K); ReadSeq(P, N, K, -1); ReadSeq(L, N, K, 0); ReadSeq(H, N, K, 0); ReadSeq(X, M, K, N); ReadSeq(Y, M, K, 0); // init tree/queries Fox1(i, N - 1) ch[P[i] - 1].pb(i); Fox(i, M) Q[X[i] - 1].pb(Y[i]); // compute answers ans = 1; nid = 0; rec(0, 0); LC[id[0]].clear(); return(ans); } int main() { int T, t; Read(T); Fox1(t, T) printf("Case #%d: %d\n", t, ProcessCase()); return(0); }