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>
 
using namespace std;
typedef long long int ll;
typedef long double ld;
#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define repr(i, n) for (ll i = (ll)(n) - 1; i >= 0; i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define INF (1e5)

#define PI (acos(-1))
#define print(x) cout << x << endl
 
ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a;}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
typedef pair <ll,ll> P;

template <typename T> 
ostream& operator<<(ostream& os, const vector<T>& v) 
{ 
    for (int i = 0; i < (int)v.size(); ++i) { 
        os << v[i]; 
        if (i != (int)v.size() - 1) 
            os << " "; 
    } 
    return os; 
} 

int main()
{
    ll M, D;
    cin >> M >> D;
    if (D < 22) print(0);
    else {
        ll cnt = 0;
        for(ll i = 22; i <= D; i++) {
            if (i / 10 < 2 || i%10 < 2) continue;
            ll tmp = (i/10) * (i%10);
            if (tmp <= M) cnt++;
            
        }
        print(cnt);
    }
    return 0;
}