/*
 * 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");
    }

}
/**
 * 
 */
package codejam2012.qualification;

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.util.HashSet;
import java.util.Set;

/**
 * @author Jaffar Ramay
 *
 *
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 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.

Limits

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

Small dataset

1 â¤ A â¤ B â¤ 1000.

Large dataset

1 â¤ A â¤ B â¤ 2000000.

Sample


Input 
 	 
4
1 9
10 40
100 500
1111 2222

Output 

Case #1: 0
Case #2: 3
Case #3: 156
Case #4: 287

Are we sure about the output to Case #4?

Yes, we're sure about the output to Case #4.
 */
public class RecycledNumbers {

	public static void solve() throws IOException {
		BufferedReader reader = new BufferedReader(new FileReader(new File("/Users/jaffar_ramay/Workspaces/eclipseWorkspace/Revision/src/codejam2012/qualification/RecycledNumbers.in")));
		BufferedWriter writer = new BufferedWriter(new FileWriter(new File("/Users/jaffar_ramay/Workspaces/eclipseWorkspace/Revision/src/codejam2012/qualification/RecycledNumbers.out")));
		String line = null;
		while((line=reader.readLine())!=null){
			int testCases = Integer.parseInt(line);
			for (int i = 1; i <= testCases; i++) {
				String[] testCase = reader.readLine().split(" ");
			     int numberA = Integer.parseInt(testCase[0]);
			     int numberB = Integer.parseInt(testCase[1]);
			     int result = getRecycledCount(numberA, numberB);
			     writer.write("Case #"+i+": "+result);
			     if(i<testCases)
			    	 writer.newLine();
			}
		}
		reader.close();
		writer.close();
	}
	public static int getRecycledCount(int numberA, int numberB) {
		int result=0;
		Set<String> set = new HashSet<String>();
		for (int i = numberA; i <= numberB; i++) {
			
			String number = i+"";
			
			for (int j = 0; j < number.length(); j++) {
				
				int firstNumber = Integer.parseInt(number);
				int index = (number.length()-(j+1));
				if(index>0 && index<number.length()){
					int secondNumber =  (Integer.parseInt(number.substring(index,number.length())+number.substring(0,index)));
					
					if(numberA<=firstNumber && firstNumber<secondNumber && secondNumber<=numberB && (firstNumber+"").length() == (secondNumber+"").length()){
						set.add(firstNumber+""+secondNumber);
					}	
				}
				
			}
			
		}
		return set.size();
	}
	
	public static void main(String[] args) throws IOException {
		solve();
	}
}
