#include<iostream>
using namespace std;
//k=p+23x=i+28y=e+33z===>23x=28y+i-p&&23x=33z+e-p
	int xyKey[23*33];
	int xzKey[23*28];
	int dealxy(int p,int i){//23x=28y+i-p
		int y=0,count=0;
		while(y<=23*33){
			if (!((28*y+i-p)%23))//x能取整数解
				xyKey[count++]=i+28*y;
			y++;
		}
		return count;//这里count是指该不定方程解得个数
	}
	int dealxz(int p,int e){
		int z=0,count=0;
		while(z<=23*28){
			if (!((33*z+e-p)%23))
				xzKey[count++]=e+33*z;
			z++;
		}
		return count;
	}
int main(){
	//freopen("1006.txt","r",stdin);
	int p,i,e,d;
	int x=0,y=0,z=0;
	int cases=0;
	while(cin>>p>>i>>e>>d,p!=-1&&i!=-1&&e!=-1&&d!=-1){
		cases++;
		if (p==i&&p==e){
			cout<<"Case "<<cases<<": the next triple peak occurs in "<<(21252-d)<<" days."<<endl;
			continue;
		}
		int lenxy=dealxy(p,i);//处理后将x有整数解时对应的K值放入数组xyKey
		int lenxz=dealxz(p,e);//处理后将y有整数解时对应的K值放入数组xzKey
		bool finded=0;
		for (int m=0;m<lenxy&&!finded;m++){
			//对数组xzKey进行二分搜索
			int key=xyKey[m];
			int low=0,high=lenxz-1;
			while(low<=high){
				int mid=(low+high)/2;
				if (xzKey[mid]==key){
					int k=key-d;
					k += 21252;
					k %= 21252;
					if( k != 0){
						cout<<"Case "<<cases<<": the next triple peak occurs in "<<k<<" days."<<endl;
						finded=1;
					break;
					}
					else 
						break;//里层循环结束
				}
				if (xzKey[mid]>key)
					high=mid-1;
				else 
					low=mid+1;
			}
		}
		if (!finded)
			cout<<"Case "<<cases<<": the next triple peak occurs in "<<"21252"<<" days."<<endl;
	}
	return 0;
}