/*
 * 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.*;

public class probC
{
	public static int N=2000001, K=0 ;
	
	public static void main (String[] args)
	{
		Scanner scan = new Scanner (System.in) ;
		
		int A[] = new int[N];
		boolean visited[] = new boolean [N];
		
		for (int i=0 ; i<A.length ; i++)
			A[i]=-1;
		
		for (int i=0 ; i<A.length ; i++)
			if (!visited[i])
				rotate(i,A,visited);
		
		int T = scan.nextInt();
		int n1, n2;
		
		for (int t=0 ; t<T ; t++)
		{
			n1 = scan.nextInt();
			n2 = scan.nextInt();
			System.out.printf("Case #%d: %d\n",(t+1),solve(n1,n2,A));
		}
	}
	public static long solve (int n1, int n2, int[] A)
	{
		int[] f = new int[K];
		long cumul[] = new long[N];
		
		int k=0 ;
		
		for (int i=n1 ; i<=n2 ; i++)
		{
			k=A[i];
			if (k!=-1)
			{
				f[k]++;
				cumul[i]=f[k]-1;
			}
			if (i!=0)
				cumul[i]+=cumul[i-1];
		}
		return cumul[n2]-cumul[n1];
	}
	
	public static int choose2 (int n)
	{
		return ((n*(n-1))/2);
	}
	public static void rotate (int n, int[] A, boolean[] visited)
	{
		visited[n]=true;
		String initial = Integer.toString(n);
		String rotated ;
		int newNum=0;
		boolean found=false;
		HashSet<Integer> set = new HashSet ();
		set.add(n); 
		
		for (int off=0  ; off<initial.length() ; off++)
		{
			rotated="";
			for (int i=0 ; i<initial.length() ; i++)
				rotated+=initial.charAt((i+off)%initial.length());
			
			newNum=Integer.valueOf(rotated);
			
			if (!set.contains(newNum) && newNum<A.length && Integer.toString(newNum).length()==initial.length())
			{
				set.add(newNum);
				found=true;
				visited[newNum]=true;
				A[newNum]=K;
				//System.out.println(n+" "+newNum);
			}
		}
		if (found)
		{
			A[n]=K;
			K++;
		}
	}
}
