/*
 * Problem C. Recycled Numbers
 * 
 * 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 
 	
Output 
 
4
1 9
10 40
100 500
1111 2222
Case #1: 0
Case #2: 3
Case #3: 156
Case #4: 287
 */

package google.codejam.y2012.qualifications;

import google.codejam.commons.Utils;

import java.io.IOException;
import java.util.StringTokenizer;

/**
 * @author Marco Lackovic <marco.lackovic@gmail.com>
 * @version 1.0, Apr 14, 2012
 */
public class RecycledNumbers {

	private static final String INPUT_FILE_NAME = "C-large.in";
	private static final String OUTPUT_FILE_NAME = "C-large.out";

	public static void main(String[] args) throws IOException {
		String[] input = Utils.readFromFile(INPUT_FILE_NAME);
		String[] output = new String[input.length];
		int x = 0;
		for (String line : input) {
			StringTokenizer st = new StringTokenizer(line);
			int a = Integer.valueOf(st.nextToken());
			int b = Integer.valueOf(st.nextToken());
			System.out.println("Computing case #" + (x + 1));
			output[x++] = "Case #" + x + ": " + recycledNumbers(a, b);
		}
		Utils.writeToFile(output, OUTPUT_FILE_NAME);
	}

	private static int recycledNumbers(int a, int b) {
		int count = 0;
		for (int n = a; n < b; n++) {
			count += getNumGreaterRotations(n, b);
		}
		return count;
	}

	private static int getNumGreaterRotations(int n, int b) {
		int count = 0;
		char[] chars = Integer.toString(n).toCharArray();
		for (int i = 0; i < chars.length; i++) {
			chars = rotateRight(chars);
			if (chars[0] != '0') {
				int m = Integer.valueOf(new String(chars));
				if (n == m) {
					break;
				} else if (n < m && m <= b) {
					count++;
				}
			}
		}
		return count;
	}

	private static char[] rotateRight(char[] chars) {
		char last = chars[chars.length - 1];
		for (int i = chars.length - 2; i >= 0; i--) {
			chars[i + 1] = chars[i];
		}
		chars[0] = last;
		return chars;
	}

}

/** QProblemC.java
 * 
 * Solves Problem C of the Qualification round of the Google Code Jam 2012
 * Tests for the 
 */

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.Scanner;

public class QProblemC {
	
	public static void main(String[] args) {
		try {
			Scanner in = new Scanner(new BufferedReader(
                    new FileReader("C-large.in")));
			BufferedWriter out = new BufferedWriter(new FileWriter(
                    new File("C-small.out")));
			int numberOfTests = in.nextInt();

            // Given integers
            int A = 0;
            int B = 0;
            
            // Variables for recycled pairs
            int n = 0;
            String nString; // To check the number of digits for equality with m
            String m = ""; // String avoids problem of losing 0 as a digit
            int mInt = 0;
            String mTemp; // Remove leading 0s before length equality
            
            // Reference for m to test when a full rotation has occurred
            String originalM = "";
            
            int count = 0;
            
			for(int i = 0; i < numberOfTests; i++) {
				System.out.print("Case #" + (i+1) + ": ");
				out.write("Case #" + (i+1) + ": ");

                A = in.nextInt();
                B = in.nextInt();
                n = A;
                nString = "" + n;
                
                // Run through all the numbers between A and B looking for
                // recycled pairs
                while(n < B) {
                    m = "" + n;
                    originalM = m; 
                    m = rotate(m);
                    
                    // Run through all possible recycles of n
                    while(!m.equals(originalM)) {
                        mInt = Integer.parseInt(m);
                        mTemp = "" + mInt; // For check without leading 0s
                        // If a recycle fits our constraints, count it
                        if(mInt > n && mInt <= B && 
                                mTemp.length() == nString.length()) {
                            count++;
                        }
                        m = rotate(m);
                    }
                    
                    n++;
                    nString = "" + n;
                }
                
                
                System.out.println(count);
                out.write(count + "\n");
                
                count = 0;
			}

			in.close();
			out.close();
		} catch (IOException e) {
			System.out.println("Error reading file. " + e);
		}
	}
    
    /** 
     * "Rotates" the given string so that the last character becomes the first
     * character.
     * 
     * @param x. The string to rotate.
     * @return String. The rotated string
     */
    public static String rotate(String x) {
        // Only take action if there are multiple characters
        if (x.length() > 1) {
            char lastChar = x.charAt(x.length() - 1);
            // Remove last char from buffer string, then add it to the front
            x = "" + lastChar + x.substring(0, x.length() - 1);
        }
        return x;
    }
}
