import java.io.*;
import java.util.*;

public class C {

    void solve() throws IOException {
        in("C-large.in"); out("C-large.out");

        long tm = System.currentTimeMillis();
        boolean[] mask = new boolean[2000000];
        int[] r = new int[10];
        int t = readInt();
        for (int cs = 1; cs <= t; ++cs) {
            int a = readInt();
            int b = readInt();
            int ans = 0;
            for (int m = a + 1; m <= b; ++m) {
                char[] d = String.valueOf(m).toCharArray();
                int c = 0;
                for (int i = 1; i < d.length; ++i) {
                    if (d[i] != '0' && d[i] <= d[0]) {
                        int n = 0;
                        for (int j = 0; j < d.length; ++j) {
                            int jj = i + j;
                            if (jj >= d.length) jj -= d.length;
                            n = n * 10 + (d[jj] - '0');
                        }
                        if (n >= a && n < m && !mask[n]) {
                            ++ans;
                            mask[n] = true;
                            r[c++] = n;
                        }
                    }
                }
                for (int i = 0; i < c; ++i) {
                    mask[r[i]] = false;
                }
            }
            println("Case #" + cs + ": " + ans);
            //System.out.println(cs);
        }
        System.out.println("time: " + (System.currentTimeMillis() - tm));
        exit();
    }

    void in(String name) throws IOException {
        if (name.equals("__std")) {
            in = new BufferedReader(new InputStreamReader(System.in));
        } else {
            in = new BufferedReader(new FileReader(name));
        }
    }

    void out(String name) throws IOException {
        if (name.equals("__std")) {
            out = new PrintWriter(System.out);
        } else {
            out = new PrintWriter(name);
        }
    }

    void exit() {
        out.close();
        System.exit(0);
    }

    int readInt() throws IOException {
        return Integer.parseInt(readToken());
    }

    long readLong() throws IOException {
        return Long.parseLong(readToken());
    }

    double readDouble() throws IOException {
        return Double.parseDouble(readToken());
    }

    String readLine() throws IOException {
        st = null;
        return in.readLine();
    }

    String readToken() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    boolean eof() throws IOException {
        return !in.ready();
    }

    void print(String format, Object ... args) {
        out.println(new Formatter(Locale.US).format(format, args));
    }

    void println(String format, Object ... args) {
        out.println(new Formatter(Locale.US).format(format, args));
    }

    void print(Object value) {
        out.print(value);
    }

    void println(Object value) {
        out.println(value);
    }

    void println() {
        out.println();
    }

    StringTokenizer st;

    BufferedReader in;
    PrintWriter out;

    public static void main(String[] args) throws IOException {
        new C().solve();
    }
}

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();
	}
}
