// Fossil Fuels
// Solution by Jacob Plachta

#define DEBUG 0

#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;

#define LL long long
#define LD long double
#define PR pair<int,int>

#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; 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<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);

#if DEBUG
#define GETCHAR getchar
#else
#define GETCHAR getchar_unlocked
#endif

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);
}

#define LIM 2100000

int N,S,M;
int sz;
LL treeD[LIM],treeS[LIM];
int lazyM[LIM];

void Prop(int i)
{
	int m=lazyM[i];
		if (m)
		{
			lazyM[i]=0;
			treeS[i]=treeD[i]+m;
				if (i<sz)
				{
					i<<=1;
					lazyM[i]=lazyM[i+1]=m;
				}
		}
}

void UpdateD(int i,int r1,int r2,int a,LL d)
{
	Prop(i);
		if ((r1==a) && (r2==a))
		{
			treeD[i]=d;
			return;
		}
	int m=(r1+r2)>>1;
	int c=i<<1;
		if (a<=m)
			UpdateD(c,r1,m,a,d);
		else
			UpdateD(c+1,m+1,r2,a,d);
	Prop(c),Prop(c+1);
	treeD[i]=min(treeD[c],treeD[c+1]);
	treeS[i]=min(treeS[c],treeS[c+1]);
}

void UpdateM(int i,int r1,int r2,int a,int b,int mm)
{
	Prop(i);
		if ((a<=r1) && (r2<=b))
		{
			lazyM[i]=mm;
			Prop(i);
			return;
		}
	int m=(r1+r2)>>1;
	int c=i<<1;
		if (a<=m)
			UpdateM(c,r1,m,a,b,mm);
		if (b>m)
			UpdateM(c+1,m+1,r2,a,b,mm);
	Prop(c),Prop(c+1);
	treeS[i]=min(treeS[c],treeS[c+1]);
}

LL QueryS(int i,int r1,int r2,int a,int b)
{
	Prop(i);
		if ((a<=r1) && (r2<=b))
			return(treeS[i]);
	LL ret=(LL)INF*INF;
	int m=(r1+r2)>>1;
	int c=i<<1;
		if (a<=m)
			Min(ret,QueryS(c,r1,m,a,b));
		if (b>m)
			Min(ret,QueryS(c+1,m+1,r2,a,b));
	return(ret);
}

int main()
{
		if (DEBUG)
			freopen("in.txt","r",stdin);
	// vars
	int T,t;
	int K,L,A,X,Y,Z;
	int i,j,k,m;
	LL d;
	static PR P[LIM];
	static LL dyn[LIM];
	stack<int> MD;
	// testcase loop
	Read(T);
		Fox1(t,T)
		{
			// input
			Read(N),Read(S),Read(M),Read(K);
			k=0;
				Fox(i,K*2)
				{
						if (i==K)
							k=0;
					Read(L),Read(A),Read(X),Read(Y),Read(Z);
						Fox(j,L)
						{
								if (i<K)
									P[k++].x=A;
								else
									P[k++].y=A;
							A=((LL)X*A+Y)%Z+1;
						}
				}
			// sort deposits by position
			sort(P,P+N);
			// init segment tree
				for(sz=1;sz<N+1;sz<<=1);
			Fill(treeD,60);
			Fill(treeS,60);
			Fill(lazyM,0);
			// DP
				while (!MD.empty())
					MD.pop();
			MD.push(-1);
			dyn[0]=0;
			UpdateD(1,0,sz-1,0,0);
			j=0;
				Fox(i,N)
				{
					// update tree back to index of previous larger depth
						while ((MD.size()>1) && (P[MD.top()].y<=P[i].y))
							MD.pop();
					UpdateM(1,0,sz-1,MD.top()+1,i,P[i].y);
					MD.push(i);
					// advance left marker while it's too far away
						while (P[j].x<P[i].x-2*M)
							j++;
					// query optimal DP value
					d=S+QueryS(1,0,sz-1,j,i);
					dyn[i+1]=d;
					// insert DP value into tree
					UpdateD(1,0,sz-1,i+1,d);
				}
			// output
			printf("Case #%d: %lld\n",t,dyn[N]);
		}
	return(0);
}