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>

int main(){
  int m,d,d1,d10,k=0,count=0;
  scanf("%d %d",&m,&d);
  int a[2][9900];
 /* d10 = d/10;
    d1 = d%10; */
  for(int i=1;i<=m;i++){//month
    for(int c=1;c<=d;c++){ //day
        d10 = c/10;
	d1 = c%10;
	if((d10<2 || d1<2)){
	}
	else if( i==d10*d1){
	a[0][k] = d10;
	a[1][k] = d1;
	k++;
	count++;
      }
    }
  }
  printf("%d\n",count);
  return 0;
}