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 <math.h>
#include <string.h>
#include <stdbool.h>  //for bool
#include <stdlib.h> // abs,labs,llabs
#include <limits.h>
#include <ctype.h>

typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld; //%Lf

int main(void) {
  int m, d, i = 0, j = 0, ans = 0, f = 0, l = 0, x = 0;
  scanf("%d %d", &m, &d);
  for (i = 1; i <= m; i++) {
    for (j = 1; j <= d; j++) {
      if (j < 22) {
        continue;
      } else {
        x = j;
        f = x % 10;
        x = x / 10;
        l = x % 10;
        if (f * l == i && f >= 2 && l >= 2) {
          ans++;
          // printf("%d %d\n", (f * l), j);
        }
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}