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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("Ofast")
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<iomanip>
#include<algorithm>
#include<vector>
#include<math.h>
#include<utility>
using namespace std;


int main(){
    int m, d;
    cin >> m >> d;

    int res = 0;
    int d1;
    int d2;

    for(int month=4;month<=m;month++){
        for(int day=22;day<=d;day++){
            int tmp = day;
            d1 = day % 10;
            if(d1 < 2) continue;
            tmp -= d1;
            d2 = tmp/10;

            if(month == d1*d2)res++;
            cout << res << endl;
        }
    }
}