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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>

int main(void){

	int a = 0;
	int b = 0;
	int c = 0;
  	int d = 0;
  	int e = 0;
  	int f = 0;
    int i = 0;
    int j = 0;

	scanf("%d", &a);
	scanf("%d", &b);
  
  	for(i = 1; i <= a; i++){
      for(j = 1; j <= b; j++){
        d = j % 10;
        e = j / 10;
        f = d * e;
        if(d >= 2 && e >= 2 && f == i){
          c++;
        }
      }
    }
	printf("%d", c);

	return 0;
}