/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

package Happy;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.util.*;
import java.util.Arrays.*;
import java.io.BufferedReader;
//import java.io.IOException;
//import java.io.InputStreamReader;
//import java.io.PrintWriter;
//import java.util.StringTokenizer;

/**
 *
 * @author ipoqi
 */
public class Happy {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        new Happy().haha();
    }

    public void haha() {
        BufferedReader in = null;
        BufferedWriter out = null;
        try{
            in = new BufferedReader(new FileReader("C-large.in"));
            out = new BufferedWriter(new FileWriter("LLL.out"));
            int T = Integer.parseInt(in.readLine());
            System.out.println("T="+T);

            //LinkedList<Integer> mm = new LinkedList<Integer>();
            //mm.add(new LinkedList<Integer>());
           //int[][] mm;
           //for(int ii=0;ii<mm.length;ii++){
           //     mm[0][ii] = 1;
           //}


            for(int i=0;i<T;i++){
                String[] line = in.readLine().split(" ");
                int A = Integer.parseInt(line[0]);
                int B = Integer.parseInt(line[1]);
                //System.out.print(" A = "+A+"\n");
                //System.out.print(" B = "+B+"\n");

                int ans = 0;

                
                for(int j=A;j<B;j++){
                    int n=j;
                    if(n>=10){
                        String N = Integer.toString(n);
                        int nlen = N.length();
                        List<Integer> oks = new ArrayList<Integer>();
                        for(int k=0;k<nlen-1;k++){
                            String M = "";
                            for(int kk=1;kk<=nlen;kk++){
                                 M = M + N.charAt((k+kk)%nlen);
                            }
                            int m = Integer.parseInt(M);
                            //System.out.print(" N = "+N+"\n");
                            //System.out.print(" M = "+M+"\n");

                            if(m>n && m<=B) {
                                boolean isNewOne = true;
                                for(int kkk=0;kkk<oks.size();kkk++){
                                    //System.out.print(" KKK = "+oks.get(kkk)+"\n");
                                    if(m==oks.get(kkk)){
                                        isNewOne = false;
                                    }
                                }
                                if(isNewOne){
                                    //System.out.print(" N = "+N+"\n");
                                    //System.out.print(" M = "+M+"\n");
                                    oks.add(m);
                                    ans++;
                                }
                            }
                        }
                    }
                }
                out.write("Case #"+(i+1)+": "+ans+"\n");
                System.out.print("Case #"+(i+1)+": "+ans+"\n");
            }
            in.close();
            out.close();
        }catch(Exception e){
            e.printStackTrace();
            try{
                in.close();
                out.close();
            }catch(Exception e1){
                e1.printStackTrace();
            }
        }
        System.out.print("YES!\n");
    }

}
/*
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-large");
        } 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);
            }
      
}
}