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 Python (3.4.3) (judge verdict: Runtime Error)

N,K = map(int,input().split())
A = [int(x) for x in input().split()]

nums = []
nums_2 = []
for i in range(N):
  num = int(A[i])
  s = 0
  k = 0
  for j in range(i,N):
    if A[j] < num:
      s += 1
  nums.append(s)
  for t in range(N):
    if A[t] < num:
      k += 1
  nums_2.append(k)

result = 0
for i in range(N):
  result += ((K*(K-1))/2)*(nums_2[i]) + K*nums[i]


print(int(result%(10E9+7)))