import java.util.Scanner;
import java.io.*;
import java.lang.Math;


public class C{
	public static int recycle(String str){
		String[] numbers = str.split(" ");
		int min = Integer.parseInt(numbers[0]);
		int max = Integer.parseInt(numbers[1]);
		int ret = 0;

		for(int i = min; i < max; i++){
			String num = i + "";
			int len = num.length();
			int k = i;
			for(int j = 1; j < len; j++){
				k = (k/10) + (int)java.lang.Math.pow(10.0,(double)(len-1))*(k%10);
				if (k > i && k <= max)
					ret++;
				else if(k == i)
					break;
			}
		}

		return ret;
	}

	public static void main(String[] args){
		Scanner sc = null;
		String next;
		int out;
		try{
			sc = new Scanner(new File("C-large.in"));
			int cases = Integer.parseInt(sc.nextLine());
			int current = 0;
			

			FileWriter fs = new FileWriter("output.txt");
			BufferedWriter buff = new BufferedWriter(fs);
			while(current < cases){
				next = sc.nextLine();
				current++;
				out = recycle(next);
				buff.write("Case #" + current + ": " + out + "\n");
			}
			buff.close();
		}
		catch(Exception ex){}
	}
}
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();
	}
}
