/*
 * 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.File;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Scanner;

public class Recycled {
    public static void main(String[] args) throws IOException {
        int[] powersOf10 = {1, 10, 100, 1000, 10000, 100000, 1000000};
        String inFileName = "C-large.in";
        Scanner in = new Scanner(new File(inFileName));
        String outFileName = inFileName.substring(0, inFileName.lastIndexOf('.')) + ".out";
        PrintWriter out = new PrintWriter(outFileName);
        int numSets = in.nextInt();
        in.nextLine();
        for(int probSet = 1; probSet <= numSets; probSet++) {
            out.print("Case #" + probSet + ": ");
            int a = in.nextInt();
            int b = in.nextInt();
            HashSet<Pair> set = new HashSet<Pair>();
            int numOne = a;
            while(numOne <= b) {
                //figure out length of a, could use logs but don't want to worry
                //about double division, rounding, ceiling, etc.
                int temp = numOne;
                int len = 1;
                while((temp /= 10) > 0) len++;
                int numTwo = numOne;
                //cycle through all possible pairs with a
                for(int i = 0; i < len - 1; i++) {
                    int dig = numTwo % 10;
                    if(dig == 0) continue; //moving a 0 from the back to the front
                                            //would change the length of the number
                    numTwo /= 10;
                    numTwo += dig * powersOf10[len - 1];
                    if(numTwo < a || numTwo > b || numOne == numTwo) continue;
                    Pair pair = new Pair(numOne, numTwo);
                    set.add(pair);
                }
                numOne++;
            }
            out.println(set.size());
        }
        in.close();
        out.close();
    }
    
    private static class Pair {
        int one, two;
        public Pair(int o, int t) {
            one = o;
            two = t;
        }

        @Override
        public int hashCode() {
            return one + two; //a bad hash, but I need something commutative
        }
        
        public boolean equals(Object obj) {
            Pair other = (Pair) obj;
            return (other.one == one && other.two == two) || 
                    (other.one == two && other.two == one);
        }
        
        public String toString() {
            return "{" + one + ", " + two + "}";
        }
    }
}
