/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

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

/**
 *
 * @author imgps
 */
public class C {
    public static void main(String args[]) throws Exception{
        int A,B;
        int ctr = 0;
        int testCases;
        
        Scanner sc = new Scanner(new FileInputStream("C-small-attempt0.in "));
        testCases = sc.nextInt();
        int input[][] = new int[testCases][2];
        for(int i=0;i<testCases;i++)
        {
          //  System.out.print("Enter input A B: ");
            input[i][0] = sc.nextInt();
            input[i][1] = sc.nextInt();
            
        }
  

for(int k=0;k<testCases;k++){
      ctr = 0;
      A = input[k][0];
      B = input[k][1];
        for(int i = A;i<=B;i++){
            int num = i;
            int nMul = 1;
            int mul = 1; 
            int tnum = num/10;
            while(tnum > 0){
                mul = mul *10;
                tnum= tnum/10;
            }
            tnum = num;
            int n = 0;
            while(tnum > 0 && mul >= 10){
                nMul = nMul * 10;
                n = (num % nMul) * mul + (num / nMul);
                mul = mul / 10;
                tnum = tnum / 10;
                for(int m=num;m<=B;m++){
                    
                    if(n == m && m > num){
                        ctr++;
                    }
                }
                
                
                
         
            }
        }
        System.out.println("Case #"+(k+1)+": "+ctr);
        
    }
    }
}

/**
 * 
 */
package google;

import hack.Tool;

import java.util.HashSet;
import java.util.List;

/**
 * http://code.google.com/codejam/contest/dashboard?c=1460488#s=p2 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.
 * 
 * @author y
 */
public class RotateNumbers {

	/**
	 * 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 ¡Ü n < m ¡Ü
	 * B?
	 * 
	 * @param min
	 * @param max
	 */
	static int pairsRotated  (int min, int max){
		if(max<10 || max<=min){return 0;}
		int ret = 0;
		
		for(int j=min; j<max; j++){
			String s = String.valueOf(j);

			HashSet<Integer> set = new HashSet<Integer>();
			
			for(int k=1; k<s.length(); k++){//iterate each digit
				
				//rotate from bigger digit
				if(s.charAt(k)>=s.charAt(0)){ //=212-> 221. ignore repeat, count (12, 21) once
					
					String rotateString = s.substring(k) + s.substring(0, k);	
					
					int rotate = Integer.valueOf(rotateString);
					
					if((j<rotate) && (rotate <=max)){ //rotate>=min
						if(!set.contains(rotate)){
							set.add(rotate);
							ret ++; 
						}						
						//System.out.println(j + "-> " + rotate);
					}
				}
			}
		}
		
		return ret;
	}

	public static void main(String[] args) {
		

		Tool.createOutput();

		List<String[]> input = Tool.readLine();

		for (int j = 0; j < input.size(); j++) {

			String[] line = input.get(j);
			int n =  Integer.parseInt(line[0]);
			int k =  Integer.parseInt(line[1]);
			int ret = pairsRotated(n, k);

			String outputLine = "Case #" + (j+1) + ": " + ret;
			System.out.println(outputLine);
			Tool.writeOutput(outputLine);
		}	


	}

}
