1006

/*
x-a = 23*a';
x-b = 28*b';
x-c = 33*c';

x === a mod 23
x === b mod 28
x === c mod 33
求x,这是中国剩余定理，
这里应用要求m1,m2,m3两两互质

x mod 23 = a mod 23
x mod 28 = b mod 28
x mod 33 = c mod 33

需要用欧拉函数了phi(n)

m = 23*28*33
M1 = m/m1 = 28*33
M2 = m/m2 = 23*33 
M3 = m/m3 = 23*28
M1' = M1^(phi(m1)-1) mod m1
M2' = M2^(phi(m2)-1) mod m2
M3' = M3^(phi(m3)-1) mod m3

x = (M1*M1'*a+M2*M2'*b+M3*M3'*c) mod m
*/

// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>

#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>

#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>

using namespace std;

// typedef
typedef __int64 LL;

// 
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)

#define Z(a,b) ((a)<<(b))
#define Y(a,b) ((a)>>(b))

const double eps = 1e-6;
const double INFf = 1e100;
const int INFi = 1000000000;
const LL INFll = (LL)1<<62;
const double Pi = acos(-1.0);

template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
	if(x>y) return x;
	return y;
}
template<class T> inline T TMIN(T x,T y)
{
	if(x<y) return x;
	return y;
}
template<class T> inline T MMAX(T x,T y,T z)
{
	return TMAX(TMAX(x,y),z);
}
template<class T> inline T MMIN(T x,T y,T z)
{
	return TMIN(TMIN(x,y),z);
}
template<class T> inline void SWAP(T &x,T &y)
{
	T t = x;
	x = y;
	y = t;
}


// code begin
int phi[50];

int a,b,c,d,M1,M2,M3,M11,M22,M33,m;
int f(int M_,int M_n,int M_mod)
{
	int ans = 1;
	for(int i=0;i<M_n;i++)
	{
		ans *= M_;
		ans %= M_mod;
	}
	return ans;
}
int main()
{
	read;
	write;
	phi[23] = 22;   // 
	phi[28] = 12;   // 28*(1-1/2)*(1-1/7) = 
	phi[33] = 20;   // 33*(1-1/3)*(1-1/11) = 20
	int ans,cas=1;
	while(scanf("%d %d %d %d",&a,&b,&c,&d)==4)
	{
		if(a==-1 && b==-1 && c==-1 && d==-1)
			break;
		a%=23;
		b%=28;
		c%=33;
		m = 23*28*33;
		M1 = m/23;
		M2 = m/28;
		M3 = m/33;
		M11 = f(M1,phi[23]-1,23);
		M22 = f(M2,phi[28]-1,28);
		M33 = f(M3,phi[33]-1,33);
		ans = (M1*M11*a+M2*M22*b+M3*M33*c)%m;
		if(ans-d<=0) ans+=m;
		printf("Case %d: the next triple peak occurs in %d days.\n",cas++,ans-d);
	}
	return 0;
}