import java.util.HashSet;
import java.util.Scanner;

public class C {

	static HashSet p = new HashSet();
	static int low;
	static int high;
	int count = 0;

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int no = sc.nextInt();

		for (int i = 1; i <= no; i++) {

			p.clear();
			low = sc.nextInt();
			high = sc.nextInt();

			for (int l = low; l <= high; l++) {
				recycle(l);
			}
			System.out.println("Case #" + i + ": " + p.size());

		}
	}

	public static void recycle(int no) {

		String s = Integer.toString(no);
		for (int i = 0; i < s.length(); i++) {
			String rec = s.substring(i) + s.substring(0, i);
			int r = Integer.parseInt(rec);
			if (r != no && r >= low && r <= high) {
				int min = Math.min(r, no);
				int max = Math.max(r, no);
				String a = Integer.toString(min) + "" + Integer.toString(max);
				p.add(a);

			}
		}

	}
}

package qualification;

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.StringWriter;

/*
 * 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 ² n < m ² 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 ² n < m ² B.

Limits

1 ² T ² 50.
A and B have the same number of digits.

Small dataset

1 ² A ² B ² 1000.

Large dataset

1 ² A ² B ² 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

Are we sure about the output to Case #4?

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

public class RecycledNumber {
	public static void main(String[] args) throws IOException{
		BufferedReader file = new BufferedReader(new FileReader(new File("src/input")));
		String firstline = file.readLine();
		int casenum = Integer.parseInt(firstline);
		StringWriter s= new StringWriter();
		for(int i=0;i<casenum;++i){
			int count = 0;
			int digit = 0;
			String temp = file.readLine();
			int index=0;
			while(temp.charAt(index)!=' '){
				++index;
				++digit;
			}
			int low = Integer.parseInt(temp.substring(0, index));
			int high = Integer.parseInt(temp.substring(index+1));
			if(1==digit){
				s.append("Case #"+(i+1)+": "+count);
				s.append("\n");
			}
			else{
				for(int j=low;j<=high;++j){
					int temp1 = j;
					int temp2 = temp1;					
					for(int k=0; k<digit-1;++k){
						int upper = temp2/10;
						int lower = temp2%10;
						if(0==lower){
							temp2 = upper;
						} else{
							temp2 = upper+lower*((int)Math.pow(10, digit-1));		
							if((temp1<temp2) && (temp2<=high))++count;
						}
					}
				}
				s.append("Case #"+(i+1)+": "+count);
				s.append("\n");
			}
		}
		FileWriter fout = new FileWriter("src/out.txt");
		fout.write(s.toString());
		fout.close();
	}
}
