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)

#include <bits/stdc++.h>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define int long long
using namespace std;
typedef pair<int, int> P;
const int mod = 1000000007;
const int INF = 1e18;

signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int m, d;
    cin >> m >> d;
    int ans = 0;
    for(int i = 1; i <= m; i++){
        for(int j = 1; j <= d; j++){
            int d2 = j / 10;
            int d1 = j % 10;
            if(d1 >= 2 && d2 >= 2 && d1 * d2 == i){
                ans++;
            }
        }
    }
    cout << ans << endl;
}