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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define bit(n,m)(((n)>>(m))&1)
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
//#define MOD 998244353
#define MOD 1000000007
#define invp(a,p)pom(a,p-2,p)

int main(){
	int a,b;
	scanf("%d%d",&a,&b);
	int ans=0;
	rep(i,22,b+1){
		int m=(i/10)*(i%10);
		if(i%10>=2&&1<=m&&m<=a)ans++;
	}
	printf("%d",ans);
}