
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;
import java.io.File;
import java.io.IOException;
import java.io.FileWriter;
import java.io.BufferedWriter;

public class Recycle
{
	public static void main(String[] args)
	{
/*		Set<Integer> perms = getPermutations(123456);
		
		for(Integer i: perms)
			System.out.println(i);*/
	
		try
		{
			Scanner s = new Scanner(new File("recycle.in"));
			BufferedWriter w = new BufferedWriter(new FileWriter(new File("recycle.out")));
			int numCases = s.nextInt();
			s.nextLine();
			
			for(int n = 1;n <= numCases; n++)
			{
				int A = s.nextInt();
				int B = s.nextInt();
				int count = 0;
				for(int k = A; k <= B; k++)
				{
					count += getNumPairs(k, B);
				}
				
				w.write("Case #" + n + ": " + count + "\n");
			}
			
			w.flush();
			w.close();
		}
		catch (IOException e)
		{
			e.printStackTrace();
		}
	}
	
	public static int getNumPairs(int n, int B)
	{
		Set<Integer> possibles = getPermutations(n);
		int count = 0;
		for(Integer i : possibles)
			if(i > n && i <= B)
				count++;
		return count;
	}
	
	public static Set<Integer> getPermutations(int n)
	{
		Set<Integer> perms = new TreeSet<Integer>();
		char[] digits = String.valueOf(n).toCharArray();
		
		for(int firstPos = 1; firstPos < digits.length; firstPos++)
		{
			String toBuild = "";
			for(int k = 0; k < digits.length; k++)
				toBuild += digits[(firstPos + k) % digits.length];
			perms.add(Integer.parseInt(toBuild));
		}
		return perms;
	}
	

}
/**
 *
 * @author Tag
 */

/*
 * Given integers A and B with the same number of digits and no leading zeros, how many distinct recycled pairs (n, m) are there with A â¤ n < m â¤ B?
 * Input
 * The first line of the input gives the number of test cases, T. T test cases follow. Each test case consists of a single line containing the integers A and B.
 * 
 * Output
 * For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1), and y is the number of recycled pairs (n, m) with A â¤ n < m â¤ B.\
 * 
 * Sample
 * Input                Output
 * 4
 * 1 9                     Case #1: 0
 * 10 40                   Case #2: 3
 * 100 500                 Case #3: 156
 * 1111 2222               Case #4: 287
 */


import java.util.Scanner;

public class RecycledNumbers 
{
    public static void main(String args[])
    {
        Scanner input = new Scanner(System.in);
        
        int numTests = new Integer(input.nextLine());
        int recycledPairs;
        
        for(int i = 1; i <= numTests; i++)
        {
            String[] tokens = input.nextLine().split(" ");
            
            Integer smallerNum = new Integer(tokens[0]);
            Integer biggerNum = new Integer(tokens[1]);
            
            recycledPairs = calcRecycled(smallerNum, biggerNum);
            
            System.out.println("Case #" + i + ": " + recycledPairs);
        }
    }

    private static int calcRecycled(Integer smallerNum, Integer biggerNum) 
    {
        int recycledPairs = 0;
        
        while(smallerNum < biggerNum)
        {
            String toCheck = smallerNum.toString();
            
            Integer tmpBigger = new Integer(biggerNum);
            
            while(smallerNum < tmpBigger)
            {
                String biggerStr = tmpBigger.toString();
                
                for(int i = 1; i < biggerStr.length(); i++)
                {
                    if((biggerStr.substring(biggerStr.length() - i) + 
                        biggerStr.substring(0, biggerStr.length() - i)).equals(toCheck))
                    {
                        recycledPairs++;
                    }
                }
                
                tmpBigger--;
            }
            
            smallerNum++;
        }
        
        return recycledPairs;
    }
}
