
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;
	}
	

}
/**
 * 
 */
package google;

import hack.Tool;

import java.util.HashSet;
import java.util.List;

/**
 * http://code.google.com/codejam/contest/dashboard?c=1460488#s=p2 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.
 * 
 * @author y
 */
public class RotateNumbers {

	/**
	 * 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 ¡Ü n < m ¡Ü
	 * B?
	 * 
	 * @param min
	 * @param max
	 */
	static int pairsRotated  (int min, int max){
		if(max<10 || max<=min){return 0;}
		int ret = 0;
		
		for(int j=min; j<max; j++){
			String s = String.valueOf(j);

			HashSet<Integer> set = new HashSet<Integer>();
			
			for(int k=1; k<s.length(); k++){//iterate each digit
				
				//rotate from bigger digit
				if(s.charAt(k)>=s.charAt(0)){ //=212-> 221. ignore repeat, count (12, 21) once
					
					String rotateString = s.substring(k) + s.substring(0, k);	
					
					int rotate = Integer.valueOf(rotateString);
					
					if((j<rotate) && (rotate <=max)){ //rotate>=min
						if(!set.contains(rotate)){
							set.add(rotate);
							ret ++; 
						}						
						//System.out.println(j + "-> " + rotate);
					}
				}
			}
		}
		
		return ret;
	}

	public static void main(String[] args) {
		

		Tool.createOutput();

		List<String[]> input = Tool.readLine();

		for (int j = 0; j < input.size(); j++) {

			String[] line = input.get(j);
			int n =  Integer.parseInt(line[0]);
			int k =  Integer.parseInt(line[1]);
			int ret = pairsRotated(n, k);

			String outputLine = "Case #" + (j+1) + ": " + ret;
			System.out.println(outputLine);
			Tool.writeOutput(outputLine);
		}	


	}

}
