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# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;

namespace pro
{
    class MainClass
    {
        public static void Main()
        {
            string[] str1 = Console.ReadLine().Split(' ');
            int n = int.Parse(str1[0]);
            long k = int.Parse(str1[1]);
            int[] a = new int[n];
            string[] str2 = Console.ReadLine().Split(' ');
            for (int i = 0; i < n; i++)
            {
                a[i] = int.Parse(str2[i]);
            }


            int count = 0;
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    if (a[i] > a[j])
                    {
                        count++;
                    }
                }
            }
            int count1 = 0;
            for (int i = 0; i < n; i++)
            {
                for (int j = i + 1; j < n; j++)
                {
                    if (a[i] > a[j])
                    {
                        count1++;
                    }
                }
            }
                long sum = 0;

            sum = (long)(count * ((k * k - k) % (long)(Math.Pow(10, 9) + 7) / 2.0));
                sum =( sum % (long)(Math.Pow(10, 9) + 7) + count1 * k % (long)(Math.Pow(10, 9) + 7) )% (long)(Math.Pow(10, 9) + 7);
                Console.WriteLine(sum);
            

        }
    }

}