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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <ctime>
#include <cstring>
#include <functional>
#include <iostream>
#include <iomanip>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

#define fix(n)          cout<<fixed<<setprecision(n)
#define rep(i,n)        for (int i = 0; i < (n); ++i)
#define sort(a)         sort((a).begin(), (a).end())
#define uniq(a)         SORT(a);(a).erase(unique((a).begin(), (a).end()), (a).end())
#define reverse(a)      reverse((a).begin(), (a).end())
#define ctos(c)         string(1, (c))
#define out(d)          cout << (d)
#define outl(d)         std::cout<<(d)<<"\n"
#define Yes()           printf("Yes\n")
#define No()            printf("No\n")
#define YES()           printf("YES\n")
#define NO()            printf("NO\n")
#define ceil(x, y)      ((x + y - 1) / (y))
#define debug(x)        cerr << #x << ": " << (x) << '\n'
#define debug2(x, y)    cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << '\n'
#define debug3(x, y, z) cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << ", " << #z << ": " << (z) << '\n'
#define dbg(v)          for (size_t _ = 0; _ < v.size(); ++_){ cerr << #v << "[" << _ << "] : " << v[_] << '\n'; }

using namespace std;
using ll = long long;
using P = pair<ll,ll>;

const ll MOD = 1000000007; // 10^9 + 7

void solve() {
  int m, d; cin >> m >> d;
  int ans = 0;
  for (int mi = 1; mi <= m; mi++) {
    for (int di = 1; di <= d; di++) {
      int d1  = di % 10;
      int d10 = di / 10;
      if (d1 >= 2 && d10 >= 2 && d1 * d10 == mi) ans++;
    }
  }
  outl(ans);
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  srand((unsigned)time(NULL));
  fix(12);

  solve();
}