/**
链接:http://acm.hdu.edu.cn/showproblem.php?pid=4135
题意:区间[l,r]与n互质的数的个数
质因数分解n,直接容斥即可
*/

#include<bits/stdc++.h>
#define ll long long
using namespace std;

/**********************************************Head-----Template****************************************/
bool Finish_read;
template<class T>inline void read(T &x){Finish_read=0;x=0;int f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;if(ch==EOF)return;ch=getchar();}while(isdigit(ch))x=x*10+ch-'0',ch=getchar();x*=f;Finish_read=1;}
template<class T>inline void print(T x){if(x/10!=0)print(x/10);putchar(x%10+'0');}
template<class T>inline void writeln(T x){if(x<0)putchar('-');x=abs(x);print(x);putchar('\n');}
template<class T>inline void write(T x){if(x<0)putchar('-');x=abs(x);print(x);}
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){ll gg=gcd(a,b);a/=gg;if(a<=LLONG_MAX/b) return a*b;return LLONG_MAX;}
/********************************Head----Temlate**********************************************/

ll l,r;
int n;
ll ans,ret;

vector<int>vec;

void dfs(int step,int flag,int tmp){
	if(step==vec.size()){
		ans+=flag*(r/tmp-(l-1)/tmp);
		return ;
	}
	dfs(step+1,flag,tmp);
	if(lcm(tmp,vec[step])<=r) dfs(step+1,-flag,lcm(tmp,vec[step]));
}

int main (){
	int t;scanf("%d",&t);
	for(int cas=1;cas<=t;cas++){
		scanf("%lld %lld %d",&l,&r,&n);
		vec.clear();ans=0;
		for(int i=2;i*i<=n;i++){
			if(n%i==0) {
				vec.push_back(i);
				while(n%i==0) n/=i;
			}
		}
		if(n>1) vec.push_back(n);
		dfs(0,1,1);
		cout<<"Case #"<<cas<<": "<<ans<<endl;
	}
	return 0;
}