費式數列推倒，矩陣作法
//============================================================================
// Name : How Many Calls.cpp
// Date : 2013/5/6 下午3:32:11
// Author : GCA
//============================================================================
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
using namespace std;
#ifdef ONLINE\_JUDGE
#define ll "%lld"
#else
#define ll "%lld"
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 10011
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define Is\_debug true
#define debug(...) if(Is\_debug)printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define FOR(it,c) for(\_\_typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
Int n,b;
Int x\[100\]\[4\];
void powmod(Int n,Int e){
if(n==1){
x\[e\]\[0\]=0;
x\[e\]\[1\]=x\[e\]\[2\]=x\[e\]\[3\]=1;
return;
}
powmod(n/2,e+1);
x\[e\]\[0\]=(x\[e+1\]\[0\]\*x\[e+1\]\[0\]+x\[e+1\]\[2\]\*x\[e+1\]\[1\])%b;
x\[e\]\[1\]=(x\[e+1\]\[1\]\*x\[e+1\]\[0\]+x\[e+1\]\[3\]\*x\[e+1\]\[1\])%b;
x\[e\]\[2\]=(x\[e+1\]\[0\]\*x\[e+1\]\[2\]+x\[e+1\]\[1\]\*x\[e+1\]\[3\])%b;
x\[e\]\[3\]=(x\[e+1\]\[1\]\*x\[e+1\]\[2\]+x\[e+1\]\[3\]\*x\[e+1\]\[3\])%b;
Int tmp\[4\];
if(n&1){
tmp\[0\]=(x\[e\]\[0\]\*x\[0\]\[0\]+x\[e\]\[2\]\*x\[0\]\[1\])%b;
tmp\[1\]=(x\[e\]\[1\]\*x\[0\]\[0\]+x\[e\]\[3\]\*x\[0\]\[1\])%b;
tmp\[2\]=(x\[e\]\[0\]\*x\[0\]\[2\]+x\[e\]\[1\]\*x\[0\]\[3\])%b;
tmp\[3\]=(x\[e\]\[1\]\*x\[0\]\[2\]+x\[e\]\[3\]\*x\[0\]\[3\])%b;
x\[e\]\[0\]=tmp\[0\];
x\[e\]\[1\]=tmp\[1\];
x\[e\]\[2\]=tmp\[2\];
x\[e\]\[3\]=tmp\[3\];
}
return;
}
int main() {
ios\_base::sync\_with\_stdio(0);
int ff=0;
x\[0\]\[0\]=0;
x\[0\]\[1\]=x\[0\]\[2\]=x\[0\]\[3\]=1;
while(~scanf("%lld%lld",&n,&b)&&b+n){
if(n!=0){
powmod(n,1);
printf("Case %d: %lld %lld %lld\\n",++ff,n,b,(2\*x\[1\]\[3\]+b-1)%b);
}
else printf("Case %d: %lld %lld 1\\n",++ff,n,b);
}
}