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: Wrong Answer)

#include <stdio.h>

int main(void){
    int m, d, d1_max, d10_max, d1, d10, cnt = 0;
    
    scanf("%d %d", &m, &d);
    
    d1_max = d % 10;
    d10_max = d / 10;
    
    for (d10 = 2; d10 < d10_max; d10++)
    {
        for (d1 = 2; d1 < 10; d1++)
        {
            if (d1 * d10 <= m)
            {
                cnt++;
            }
            else
            {
                break;
            }
        }
    }
    
    if (d10_max >= 2)
    {
        for (d1 = 2; d1 < d1_max; d1++)
        {
            if (d1 * d10_max <= m)
            {
                cnt++;
            }
            else
            {
                break;
            }
        }
    }
    
    printf("%d\n", cnt);
    
    return 0;
}