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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
using P   = pair<ll, ll>;
using mat = vector<vector<ll>>;
typedef string::const_iterator State;
class ParseError {};

void IOS() { ios::sync_with_stdio(false), cin.tie(0); }

constexpr ll INF = 1e16;
constexpr ll INM = 114514;
constexpr ll MOD = 1e9 + 7;
constexpr ld EPS = 1e-12;

template <typename T>
void dump(T x) { cout << x << endl; }
void dumpf(ld x, int t) { cout << setprecision(t) << fixed << x << endl; }
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
	fill((T *)array, (T *)(array + N), val);
}
inline ll mod(ll a, ll b) {
	if (a > 0) return a % b;
	if (a % b == 0) return 0;
	ll x = -a / b + 1;
	a += x * b;
	return a % b;
}
ll powm(ll a, ll b, ll c) {
	if (b == 0)
		return 1;
	else if (b % 2 == 0) {
		ll d = powm(a, b / 2, c);
		return (d * d) % c;
	} else
		return (a * powm(a, b - 1, c)) % c;
}
ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll dx[4] = {1, 0, -1, 0};
ll dy[4] = {0, 1, 0, -1};

int main() {
	IOS();
	ll m, d;
	cin >> m >> d;
	ll cnt = 0;
	for (int i = 1; i <= m; i++) {
		for (int j = 1; j <= d; j++) {
			int d1  = j % 10;
			int d10 = j / 10;
			if (d1 >= 2 && d10 >= 2 && d1 * d10 == i) cnt++;
		}
	}
	dump(cnt);
	return 0;
}