#include <iostream>  
using namespace std;  
int exgcd(int a,int b,int &x,int &y)  
{  
    if(a==0)  
    {  
        x=0;  
        y=1;  
        return b;  
    }  
    int g = exgcd(b%a,a,x,y);  
    int tem = y;  
    y=x;  
    x=tem-(b/a)*y;  
    return g;  
}  
int inv(int a,int n)//求逆元  
{  
    int x,y;  
    exgcd(a,n,x,y);  
    return (x%n+n)%n;  
}  
int main()  
{  
    int N,P,E,I,D,ans;  
    const int m1=23,m2=28,m3=33,M1=28*33,M2=23*33,M3=23*28,m=23*28*33;  
    const int M11 = inv(M1,m1),M22 = inv(M2,m2),M33 = inv(M3,m3);//求mi的逆元   
    scanf("%d",&N);  
    while(scanf("%d%d%d%d",&P,&E,&I,&D)!=EOF)  
    {  
        if(P==-1&&E==-1&&I==-1&&D==-1)break;  
        //a[i]*(mi*mi(逆)mod ni)因为在求逆元的时候已经mod ni了   
        //(mi*mi(逆)mod ni)是ci   
        ans = (P*M1*M11 + E*M2*M22 + I*M3*M33)%m;  
        ans -= D;  
        if(ans<=0) ans+=m;  
        printf("Case %d: the next triple peak occurs in %d days.\n",N++,ans);  
        //cout<<"Case "<<N++<<": the next triple peak occurs in "<<i<<" days.\n";  
    }  
    return 0;  
}