import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;

public class Recycled {
    public static void main(String[] args) {
        try {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            int t = Integer.parseInt(br.readLine());
            for (int i = 0; i < t; i++) {
                String[] str = br.readLine().split(" ");
                int a = Integer.parseInt(str[0]);
                int b = Integer.parseInt(str[1]);
                System.out.println("Case #" + (i + 1) + ": " + find(a, b));
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    private static int find(int a, int b) {
        HashSet<Long> used = new HashSet<Long>();
        int digits = String.valueOf(a).length();
        if (digits == 1) return 0;
        for (int i = a; i <= b; i++) {
            int s = 10;
            int m = (int) Math.pow(10, digits-1);
            for (int j = 0; j < digits-1; j++) {
                int r = i % s;
                if (r < s/10) continue;
                int q = i / s;
                int rn = r * m + q;
                s *= 10; m /= 10;
                if (i != rn && rn >= a && rn <= b) {
                    long pp;
                    if (rn < i) pp = (long)rn << 30 | i;
                    else pp = (long)i << 30 | rn;
                    if (!used.contains(pp)) {
                        used.add(pp);
                    }
                }
            }
        }
        return used.size();
    }
}

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();
	}
}
