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: Accepted)

#pragma comment(linker, "/stack:247474112")
#pragma GCC optimize("Ofast")
#include"bits/stdc++.h"
using namespace std;
#define IOS ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define int long long
#define ends ' '
#define endl '\n'
#define pb emplace_back
#define x first
#define y second
#define pii pair<int, int>
#define sz(x) (int)(x).size()

double PI = acos(-1.0);

int32_t main()
{
	// #ifndef ONLINE_JUDGE
	// 	freopen("input.txt", "r", stdin);
	// 	freopen("output.txt", "w", stdout);
	// #endif
	IOS;

	int m, d;
	cin >> m >> d;

	int ans(0);

	for(int i = 1; i <= m; ++i)
	{
		for(int j = 1; j <= d; ++j)
		{
			if(j >= 10 and j%10 >= 2 and (j/10) >= 2 and ((j%10)*(j/10) == i))
			{
				// cout << i << ' ' << j << '\n';
				++ans;
			}
		}
	}
	cout << ans;

	return 0;
}