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)

// template
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<long long, long long> LP;
#define INF 999999999
#define MOD 1000000007
#define REP(i, n) for(ll i = 0, i##_len = (n); i < i##_len; i++)
#define REP_AB(i, a, b) for (ll i = ll(a); i < ll(b); i++)
#define ALL(v) v.begin(), v.end()
#define SORT(v) sort(ALL(v))
#define UNIQUE(v) sort(ALL(v));v.erase(unique(ALL(v)), v.end());
#define SIZE(x) ((ll)(x).size())
#define REVERSE(v) reverse(ALL(v))
// true / false
#define BIN_SEARCH(v, a) binary_search(ALL(v), a)
// index
#define BIN_LEFT(v, a) (lower_bound(ALL(v), a) - v.begin())
#define BIN_RIGHT(v, a) (upper_bound(ALL(v), a) - v.begin())
#define BIN_INSERT(v, a) (v.insert(v.begin() + BIN_LEFT(v, a), a))
#define DEL(v, i) v.erase(v.begin() + i)
#define INSERT(v, i, a) v.insert(v.begin() + i, a)
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;}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
#define out cout
#define in cin
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
// template end
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    ll M, D, ans = 0;
    in >> M >> D;

    REP_AB(i, 22, D+1){
        if( (M >= (i%10)*(i/10)) && (i%10 > 1) ){
            ans++;
        } 
    }
    out << ans << endl;

    
 
    //vector<ll> X(N);

}