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));
			
		}
		
		
		

	}
	
	
}
/**
 *
 * @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;
    }
}
