Problem:
Score : 200 points

Problem Statement

Today is August 24, one of the five Product Days in a year.

A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day):

d_1 \geq 2

d_{10} \geq 2

d_1 \times d_{10} = m

Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D.

In Takahashi Calendar, how many Product Days does a year have?

Constraints

All values in input are integers.

1 \leq M \leq 100

1 \leq D \leq 99

Input

Input is given from Standard Input in the following format:

M D

Output

Print the number of Product Days in a year in Takahashi Calender.

Sample Input 1

15 40

Sample Output 1

10

There are 10 Product Days in a year, as follows (m-d denotes Month m, Day d):

4-22

6-23

6-32

8-24

9-33

10-25

12-26

12-34

14-27

15-35

Sample Input 2

12 31

Sample Output 2

5

Sample Input 3

1 1

Sample Output 3

0

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define ll long long
#define rep(i,l,r,d) for(ll i=(l);i<(r);i+=(d))
#define max(p,q) ((p)>(q)?(p):(q))
#define min(p,q) ((p)<(q)?(p):(q))
#define abs(p) ((p)>=(0)?(p):(-(p)))
int main(void){
    ll M, D, ANS=0, d1, d10;
    scanf("%lld %lld", &M, &D);
    rep(i, 4, M+1, 1){
        rep(j, 22, D+1, 1){
            d1 = j%10;
            if(d1!=1){
                d10 = j/10;
                if(i==d1*d10)ANS++;
            }
        }
    }
    printf("%lld\n", ANS);
    return 0;
}