import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

class RecycledNumbers{
	
	
	
	int getcount(int small,int large){
		int count = 0;
		for(int currnum = small; currnum < large; currnum++){
			int permut = currnum;
			int no_of_digits = 0;
			while(permut > 0){
				permut = permut/10;
				no_of_digits++;
			}
			int lastDigit = currnum % 10;
			permut = currnum / 10;
			permut = permut + lastDigit * (int)Math.pow(10, no_of_digits-1);
			while(permut != currnum){
				
				if(permut >= currnum && permut <= large){
					
					count++;
				}
				lastDigit = permut % 10;
				permut = permut / 10;
				permut = permut + lastDigit * (int)Math.pow(10, no_of_digits-1);
				
			}
			
		}
		
		return count;
	}
	
	public static void main(String args[]){
		RecycledNumbers d = new RecycledNumbers();
		
		int no_of_cases = 0;
		String input = "";
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		try{
			no_of_cases = Integer.parseInt(br.readLine());
		}
		catch(Exception e){
			System.out.println("First line is not a number");
		}
		
		for(int i=1; i <= no_of_cases;i++){
			try{
				input = br.readLine();
			}
			catch(Exception e){
				System.out.println("Number of test cases different from the number mentioned in first line");
			}
			
			System.out.print("Case #" +i+": ");
			
			StringTokenizer t =  new StringTokenizer(input);
			int small=0,large=0;
			if(t.hasMoreTokens()){
				small = Integer.parseInt(t.nextToken());
			}
			if(t.hasMoreTokens()){
				large = Integer.parseInt(t.nextToken());
			}
			System.out.println(d.getcount(small,large));
			
		}
		
		
		

	}
	
	
}
/*
 *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?
 *
 **/

import java.io.*;
public class Recycled_Numbers
{
	public static void main(String[] args)
	{
		try
		{
			BufferedReader br=new BufferedReader(new InputStreamReader(new FileInputStream(new File("g:/H.txt"))));
			int n=Integer.parseInt(br.readLine());
			String str=null;
			int A,B;
			String arr[]=new String[n];
			String array[];
			File f=new File("g:/OutputX.txt");
			f.createNewFile();
			int count=0;
			int temp1,temp2;
			boolean result=false;
			for(int i=0;i<n;i++)
			{
				str=br.readLine();
				count=0;
				array=str.split(" ");
				A=Integer.parseInt(array[0]);
				B=Integer.parseInt(array[1]);
				temp1=A;
				for(;temp1<B;temp1++)
				{
					for(int j=temp1+1;j<=B;j++)
					{
						result=getResult(temp1,j);
						if(result)
							 count++;
					}
				}
				arr[i]=new String("Case #"+(i+1)+": "+count);
			}
			FileWriter bout=new FileWriter(f);
			for(int i=0;i<n;i++)
			{
				bout.append(arr[i]+"\n");			
			}
			bout.close();
		}
		
		catch(IOException e1)
		{
			e1.printStackTrace();
		}	
		catch(Exception e)
		{
			e.printStackTrace();
		}	
	}
	public static boolean getResult(int A,int B)
	{
		boolean flag=true;
		String str=B+"";
		String str1=null;
		int length=str.length();
		//System.out.println(str);
		if(length==1)
		{
			if(A==B)
				return true;
			else 
				return false;
		}
		int count=0;
		//System.out.println("B Number"+B);
		while(A!=B&&flag)
		{
			count++;
			//System.out.println("Count"+count);
			//System.out.println("Length"+length);
			//System.out.println("last " +str.substring(length-count,length) + "first" +str.substring(0,length-count+1));
		    str1=str.substring(length-count,length)+str.substring(0,length-count);
		    //System.out.println("new Number " +str1);
		    B=Integer.parseInt(str1);	
		    if(A==B)
		    	return true;
		    if(length==count+1)
		    	flag=false;		
		    		
		}
		return false;
	}
}