/*
 * 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;
	}

}


import java.io.BufferedReader;
import java.io.FileReader;
import java.util.ArrayList;

/**
 * Problem C
 *
 * @author thomas
 */
public class RecycledNumbers {

    public static void main(String[] args) {
        String linea;
        try {
            FileReader fis = new FileReader("C-large.in");
            BufferedReader buffer = new BufferedReader(fis);
            linea = buffer.readLine();
            int T = Integer.parseInt(linea.trim());
            int A, B;
            String[] aux;
            for (int caso = 1; caso <= T; caso++) {
                linea = buffer.readLine();
                aux = linea.split(" ");
                A = Integer.parseInt(aux[0]);
                B = Integer.parseInt(aux[1]);
                procesar(caso, A, B);
            }
            buffer.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public static void procesar(int caso, int A, int B) {
//        System.out.println("Case #" + caso + ": " + A + " " + B);
        int numPair = 0;
        for (int N = A; N <= B; N++) {
            numPair += pairOf(N, B);
        }
        System.out.println("Case #" + caso + ": " + numPair);
    }

    public static void main_p(String[] args) {
//        pairOf(12345);
//        pairOf(101);
    }
    
    public static int pairOf(int N, int B) {
        ArrayList<Integer> pares = new ArrayList<Integer>();
        String sN = N + "";
        //Si es un solo digito no hay par
        if (sN.length() <= 1) {
            return 0;
        }

        //Evaluamos si el numero tiene los digitos iguales
        //En este caso el par es el mismo numero
        //Y como los pares no pueden ser iguales se descarta este caso
        
        boolean sameDigit = true;
        for (int i = 1; i < sN.length() && sameDigit; i++) {
            sameDigit = sN.charAt(i - 1) == sN.charAt(i);
        }
        if (sameDigit) {
            return 0;
        }


        //Comenzamos a buscar los pares
        String N1, N2;
        int split = 1;
        int M, numPair = 0;
        while (split < sN.length()) {
            N1 = sN.substring(0, split);
            N2 = sN.substring(split);
            M = Integer.parseInt(N2 + N1);
            //Si empieza en cero, ya los digitos no concuerdan
            //O Si M es mayor que B se sale del rango
            //O si N>=M no cumple con la restriccion A â¤ n < m â¤ B 
            if ((N2.indexOf("0") == 0) || (M > B) || (N>=M)) {
                split++;
                continue;
            }
            if(!pares.contains(M)){
                pares.add(M);
            }else{
                split++;
                continue;
            }
            numPair++;
            debug(sN + " => " + M + "[" + N1 + "-" + N2 + "]");

            split++;
        }
        return numPair;
    }

    public static void debug(Object obj) {
//        System.out.println(obj);
    }
}
