
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;
import java.io.File;
import java.io.IOException;
import java.io.FileWriter;
import java.io.BufferedWriter;

public class Recycle
{
	public static void main(String[] args)
	{
/*		Set<Integer> perms = getPermutations(123456);
		
		for(Integer i: perms)
			System.out.println(i);*/
	
		try
		{
			Scanner s = new Scanner(new File("recycle.in"));
			BufferedWriter w = new BufferedWriter(new FileWriter(new File("recycle.out")));
			int numCases = s.nextInt();
			s.nextLine();
			
			for(int n = 1;n <= numCases; n++)
			{
				int A = s.nextInt();
				int B = s.nextInt();
				int count = 0;
				for(int k = A; k <= B; k++)
				{
					count += getNumPairs(k, B);
				}
				
				w.write("Case #" + n + ": " + count + "\n");
			}
			
			w.flush();
			w.close();
		}
		catch (IOException e)
		{
			e.printStackTrace();
		}
	}
	
	public static int getNumPairs(int n, int B)
	{
		Set<Integer> possibles = getPermutations(n);
		int count = 0;
		for(Integer i : possibles)
			if(i > n && i <= B)
				count++;
		return count;
	}
	
	public static Set<Integer> getPermutations(int n)
	{
		Set<Integer> perms = new TreeSet<Integer>();
		char[] digits = String.valueOf(n).toCharArray();
		
		for(int firstPos = 1; firstPos < digits.length; firstPos++)
		{
			String toBuild = "";
			for(int k = 0; k < digits.length; k++)
				toBuild += digits[(firstPos + k) % digits.length];
			perms.add(Integer.parseInt(toBuild));
		}
		return perms;
	}
	

}
package qC;

import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;

import qB.B.Pair;

//Letfs 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 (excluding leading zeros) in order to be a recycled pair.

//Given integers A and B with the same number of digits, how many distinct
//recycled pairs (n, m) are there with A ? n < m ? B?

public class C2 {

	static int solve(int number, int b) {
		int numDigi = number;
		int orgNumber = number;
		int c = 1;
		int p = 1;
		int counter = 0;

		do {
			numDigi = numDigi / 10;
			p = p * 10;
			c++;
		} while (numDigi != 0);
		int k = c - 1;
		p = p / 10;

		for ( int i = 1; i <= k; i++) {
			number = number % p * 10 + number / p;
			if (number > orgNumber && number <= b) {
				// System.out.println( "no: " + number + ", " + check);
				counter++;
			}
		}
		return counter;
	}

	public static void main(String[] args) throws FileNotFoundException {
		if (args.length == 0) {
			throw new IllegalArgumentException("no file given");
		}
		String filename = args[0];
		Scanner scanner = new Scanner(new File(filename));
		int numOfLines = scanner.nextInt();
		scanner.nextLine();

		for (int lineIdx = 0; lineIdx < numOfLines; lineIdx++) {
			final int start = scanner.nextInt();
			final int end = scanner.nextInt();
			int ans = 0;
			for (int l = start; l <= end; l++) {
				ans += solve(l, end);
			}
			System.out.printf("Case #%d: %d\n", lineIdx + 1, ans);
		}
	}
}
