
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;
	}
	

}
/*
Problem

Do you ever become frustrated with television because you keep seeing the same things, recycled over and over again? Well I personally don't care about television, but I do sometimes feel that way about numbers.

Let's say a pair of distinct positive integers (n, m) is recycled if you can obtain m by moving some digits from the back of n to the front without changing their order. For example, (12345, 34512) is a recycled pair since you can obtain 34512 by moving 345 from the end of 12345 to the front. Note that n and m must have the same number of digits in order to be a recycled pair. Neither n nor m can have leading zeros.

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 = n < m = 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 = n < m = B.

Limits

1 = T = 50.
A and B have the same number of digits.

Small dataset

1 = A = B = 1000.

Large dataset

1 = A = B = 2000000.

Sample


Input 
 	
Output 
 
4
1 9
10 40
100 500
1111 2222
Case #1: 0
Case #2: 3
Case #3: 156
Case #4: 287

*/

import java.io.*;
import java.util.*;
import java.util.logging.Level;
import java.util.logging.Logger;
public class RecycledNumbers

{
     BufferedReader read;
     BufferedWriter write;
    public static void main(String args[])
    {
        try {
            new RecycledNumbers().init("C-small-attempt0");
        } catch (Exception ex) {
            //Logger.getLogger(RecycledNumbers.class.getName()).log(Level.SEVERE, null, ex);
			System.out.println("Error" + ex);
			
        }
    }
     void init(String name) throws Exception
    {
        read=new BufferedReader(new FileReader(new File(name+".in")));
        write=new BufferedWriter(new FileWriter(new File(name+".out")));
        
		String x="";
        
            try {
                   x = read.readLine();
                   int nos = Integer.parseInt(x.trim());
                   for(int z=0;z<nos;z++)
                      {
					      
                                  x = read.readLine();
								  String ss[]=x.split(" ");
								 // int[] anb = new int[ss.length];
								   int a =  Integer.parseInt(ss[0].trim());
								   int b =  Integer.parseInt(ss[1].trim());;
								   //System.out.println(a+"  "+b );
								   //a=10;b=40;
									  int count=0;
								   for(int i=a;i<=b;i++)
								    {
									  String sub="",s1="";
									  String s= "" + i;
									  //System.out.println("....   "+s);
									  int len = s.length();
									  ArrayList<Integer> ar=new ArrayList<Integer>();
									  for(int j=0;j<len;j++)
										{
									      String c = ""+(s.charAt(len-1));
										  // System.out.print("....   "+c);
										  sub = s.substring(0,len-1);
										   //System.out.print("....   "+sub);
										//  s1 = c + sub;
										  s = c + sub;
										 //System.out.print("....   "+s);
										  if(s.charAt(0)!='0')
										   {
										     int num = Integer.parseInt(s);
											 if(!ar.contains(num)&&num>i && num<=b)
											 {
											 ar.add(num);
											 //System.out.println("   i....   "+i+"...num....   "+num);
											 count++;}
										   }
									    }
									}

								write.write("Case #"+(z+1)+":"+" "+count+"\n");
							System.out.println("Case #"+(z+1)+":"+" "+count+"\n");
					 }                
				write.flush();
                   write.close();
                   read.close();
                  // System.out.println(a+"    "+b);
                
              }
             catch (Exception ex) {
			 System.out.println("Error" + ex);
                           System.exit(0);
            }
      
}
}