/*
 * 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.util.ArrayList;
import java.util.LinkedList;

public class RecycledNumbers {
	static String[] zeros = { "0", "00", "000", "0000", "00000", "000000" };

	public static void solve(String fname) {
		Read r = new Read();
		r.Read(fname);
		int numOfTestCases = r.noOfTestCases;
		int[] numbers = r.nums;
		String finalAnser = "";
		for (int i = 0; i < numOfTestCases; i++) {
			String answer = "";
			int firstNum = numbers[2 * i];
			int secondNum = numbers[2 * i + 1];
			
			int lastNumIndex = (int) (secondNum / Math.pow(10, ((secondNum+"").length())-1));
			lastNumIndex++;
			String large="";
			for (int k = 9; k >=lastNumIndex; k--) {
				//check large number
				
					large=large+" "+k;
				
				
			}
			// int mid = ((secondNum-firstNum)/2)+1;
			int noOfans = 0;
			for (int j = firstNum; j < secondNum; j++) {
				String numb = j + "";
				int firstIndex = -1;
				int lastIndex = -1;

				for (int k = zeros.length - 1; k >= 0; k--) {
					// check zeros 
					if (numb.contains(zeros[k])) {
						firstIndex = numb.indexOf(zeros[k]);
						//first index of 0 eg in 201000 it will be 3
						lastIndex = firstIndex + zeros[k].length() - 1;
						//last index = 3+(3-1) = 5 
					}
				}
			
			

				// /LinkedList<Integer> number = new LinkedList<Integer>();
				for (int k = 0; (k < numb.length()) ; k++) {
					if(k>=firstIndex&& (k <= lastIndex)&&(!large.contains(numb.charAt(k)+""))){
					//	System.out.println(k);
					}else{
					String first = numb.substring(0, k);
					String last = numb.substring(k, numb.length());
				//	int la = Integer.parseInt(last);
					String newNumb = last + first;
				//	int biggest = (int) (la * Math.pow(10, numb.length()- last.length()));
		
						int numbC = Integer.parseInt(newNumb);
						if ((numbC <= secondNum) && (numbC > j)) {
							noOfans++;
						}
					
				}
				}
				int x = 0;
				// int k=0;
				// while(!number.isEmpty()){
				// x=(int) (x+number.removeLast()*Math.pow(10, k));
				// k++;
				// }

			}
			
			answer = "Case #" + (i + 1) + ": " + noOfans;
		//	System.out.println(answer);
			finalAnser = finalAnser + answer + "\n";

		}
		r.write("out", finalAnser);
		System.out.println(finalAnser);
	}
	public static void main(String[] args) {

		solve("test");
		// String n = "10002";
		// System.out.println(n.lastIndexOf('0'));
	}
}
