/* Check cf5-opt.vim defs.
VIM: let g:lcppflags="-std=c++11 -O2 -pthread"
VIM: let g:wcppflags="/O2 /EHsc /DWIN32"
VIM: let g:cppflags=g:Iboost.g:Itbb
VIM: let g:ldflags=g:Lboost.g:Ltbb.g:tbbmalloc.g:tbbmproxy
VIM: let g:ldlibpath=g:Bboost.g:Btbb
VIM: let g:argv=""
*/

/*
New Lottery Game

The Lottery is changing! The Lottery used to have a machine to generate a random winning number. But due to cheating problems, the Lottery has decided to add another machine. The new winning number will be the result of the bitwise-AND operation between the two random numbers generated by the two machines.

To find the bitwise-AND of X and Y, write them both in binary; then a bit in the result in binary has a 1 if the corresponding bits of X and Y were both 1, and a 0 otherwise. In most programming languages, the bitwise-AND of X and Y is written X&Y.

For example:
   The old machine generates the number 7 = 0111.
   The new machine generates the number 11 = 1011.
   The winning number will be (7 AND 11) = (0111 AND 1011) = 0011 = 3.

With this measure, the Lottery expects to reduce the cases of fraudulent claims, but unfortunately an employee from the Lottery company has leaked the following information: the old machine will always generate a non-negative integer less than A and the new one will always generate a non-negative integer less than B. 

Catalina wants to win this lottery and to give it a try she decided to buy all non-negative integers less than K. 

Given A, B and K, Catalina would like to know in how many different ways the machines can generate a pair of numbers that will make her a winner. 

Could you help her?

Input

The first line of the input gives the number of test cases, T. T lines follow, each line with three numbers A B K.

Output

For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1) and y is the number of possible pairs that the machines can generate to make Catalina a winner.

Limits

1 ? T ? 100.
Small dataset

1 ? A ? 1000.
1 ? B ? 1000.
1 ? K ? 1000.
Large dataset

1 ? A ? 109.
1 ? B ? 109.
1 ? K ? 109.
Sample


Input 
 	
Output 
 
5
3 4 2
4 5 2
7 8 5
45 56 35
103 143 88

Case #1: 10
Case #2: 16
Case #3: 52
Case #4: 2411
Case #5: 14377
In the first test case, these are the 10 possible pairs generated by the old and new machine respectively that will make her a winner: <0,0>, <0,1>, <0,2>, <0,3>, <1,0>, <1,1>, <1,2>, <1,3>, <2,0> and <2,1>. Notice that <0,1> is not the same as <1,0>. Also, although the pair <2, 2> could be generated by the machines it wouldn't make Catalina win since (2 AND 2) = 2 and she only bought the numbers 0 and 1.
*/

#include <assert.h>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <exception>
#include <stdexcept>
#include <map>
#include <set>
#include <list>
#include <vector>
#include <string>
#include <memory>
#include <functional>
#include <algorithm>
#include <utility>
#include <limits>
#include <math.h>

typedef long long i64;
typedef std::vector<i64> vec;
void check( bool b ) { 	if ( !b ) 
	std::cerr << "error" << std::endl; }
#define FOR(i,l) for ( i64 i =0, ie = i64(l); i<ie; ++i )
/*
	Read n values into v
*/
template <class V>
void readv( V& v, int n )
{
	v.reserve(n);
	for ( int i = 0; i < n; ++i )
	{
		typename V::value_type e;
		std::cin >> e;
		check( !std::cin.fail() );
		v.push_back(e);
	}
}


int solve_puzzle()
{
	int a, b, k;
	std::cin >> a >> b >> k;

	i64 s = 0;
	for ( int i = 0; i < k; ++i )
	for ( int i = 0; i < a; i++ )
		for ( int j = 0; j < b; j++ )
			if ( (i&j) < k )
				s++;

	return s;
}

#undef int
int main ( void )
{try{
	srand((unsigned)time(NULL));
	int puzzle_count;

	std::cin >>	puzzle_count;
	std::cin.ignore(std::numeric_limits<std::streamsize>::max(),'\n');
	for ( int i = 1; i <= puzzle_count; i++ )
	{
		std::cout << "Case #" << i << ": ";
		auto r = solve_puzzle();
		std::cout << r << std::endl;
	}

	return 0;
}
catch ( const std::exception& e )
{
	std::cerr << std::endl
			<< "std::exception(\"" << e.what() << "\")." << std::endl;
	return 2;
}
catch ( ... )
{
	std::cerr  << std::endl
			<< "unknown exception." << std::endl;
	return 1;
}}
