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: Wrong Answer)

#include <cstdio>
#include <iostream>
#include <string>
#include <vector>
#include <array>
#include <algorithm>
#include <utility>
#include <numeric>
#include <deque>
#include <stack>
#include <queue>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <cmath>
#include <limits>

#define for0(i, n)    for (int i = 0  ; i <  (n); ++i)
#define for1n(i, n)   for (int i = 1  ; i <= (n); ++i)
#define for1(i, n)    for (int i = 1  ; i <  (n); ++i)
#define forn(i, n)    for (int i = 0  ; i <= (n); ++i)
#define forx(i, n, x) for (int i = (x); i <  (n); ++i)

using namespace std;

typedef long long ll;
typedef priority_queue<ll> pq;
typedef priority_queue<ll,vector<ll> ,greater<ll>> pql;
typedef stack<ll> stk;
typedef queue<ll> qu;


int main(void){
    int M, D;
    cin >> M >> D;
    int result = 0;

    for1n(i, M){
        for1n(j, D){
            int d1 = j%10;
            int d10 = j/10;
            if (d1>=2 && d10==2 && i == d1 * d10) {
                ++result;
            }
        }
    }

    cout << result;

    return 0;
}