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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc *bufio.Scanner

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func main() {
	sc = bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	m, d := nextInt(), nextInt()

	ans := 0
	for mth := 1; mth <= m; mth++ {
		for div := 2; div <= 9; div++ {
			if mth/div > 1 && mth%div == 0 {
				if (mth/div)*10+div <= d {
					ans++
				}
				if div*10+(mth/div) <= d {
					ans++
				}
			}
		}
	}
	fmt.Println(ans / 2)
}