import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;

public class Recycled {
    public static void main(String[] args) {
        try {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            int t = Integer.parseInt(br.readLine());
            for (int i = 0; i < t; i++) {
                String[] str = br.readLine().split(" ");
                int a = Integer.parseInt(str[0]);
                int b = Integer.parseInt(str[1]);
                System.out.println("Case #" + (i + 1) + ": " + find(a, b));
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    private static int find(int a, int b) {
        HashSet<Long> used = new HashSet<Long>();
        int digits = String.valueOf(a).length();
        if (digits == 1) return 0;
        for (int i = a; i <= b; i++) {
            int s = 10;
            int m = (int) Math.pow(10, digits-1);
            for (int j = 0; j < digits-1; j++) {
                int r = i % s;
                if (r < s/10) continue;
                int q = i / s;
                int rn = r * m + q;
                s *= 10; m /= 10;
                if (i != rn && rn >= a && rn <= b) {
                    long pp;
                    if (rn < i) pp = (long)rn << 30 | i;
                    else pp = (long)i << 30 | rn;
                    if (!used.contains(pp)) {
                        used.add(pp);
                    }
                }
            }
        }
        return used.size();
    }
}

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);
		}
	}
}
