#include <iostream>
using namespace std;
#define maxn 70
long long prime[maxn];
long long getans(long long num,int m) //求1-mun 不互素的数，m为num素因子的个数；
{ //利用容斥原理
long long ans=0,tmp,i,j,flag;
for(i=1;i<(long long)(1<<m);i++){
tmp=1,flag=0;
for(j=0;j<m;j++)
if(i&((long long)(1<<j)))
flag++,tmp*=prime[j];
if(flag&1)
ans+=num/tmp;
else
ans-=num/tmp;
}
return ans;
}
int main()
{
int T,t=0,m;
long long n,a,b;
cin>>T;
while(T--){
cin>>a>>b>>n;
m=0;
for(int i=2;i*i<n;i++){
if(n&&n%i==0){
prime[m++]=i;
while(n&&n%i==0)
n/=i;
}
}
if(n>1)prime[m++]=n;
cout<<"Case #"<<++t<<": "<<(b-getans(b,m))-(a-1-getans(a-1,m))<<endl;
}
return 0;
}