//每次枚举前四位,把前四位反转拼接到后面去,这样就是在回文数里判断一个数是不是合法日期

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;public class Main{static int n,m,res;static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));//这里要从0开始,才会跟月份对应static int[] days = {0,31,28,31,30,31,30,31,31,30,31,30,31};public static void main(String[] args)throws IOException {n = Integer.parseInt(in.readLine());m = Integer.parseInt(in.readLine());//只枚举前四位for (int i = 1000; i < 9999; i++) {int x = i,t = i;//将前四位反转拼接for (int j = 0; j < 4; j++) {x = x * 10 + t % 10;t /= 10;}//合法日期并且符合题意就算一个答案if (x >= n && x <= m && check(x)) res++;}System.out.println(res);in.close();}//判断日期合法public static boolean check(int date){int year = date / 10000;    //丢掉后面四位int month = date / 100 % 100;   //丢掉后面两位拿到后面两位int day = date % 100;   //拿到后面两位//先判断月,再判断日,判断日的时候先不判断2月,2月单独判断if (month == 0 || month > 12) return false;if (day == 0 || (month != 2 &&day > days[month]))return false;//月份为2单独判断if (month == 2){int etra = 0;if (year % 400 == 0 || year % 4 == 0 && year % 100 != 0) etra++;if (day > 28 + etra) return false;}return true;}
}