Problem:
Max Score: $250$ Points

Problem Statement

Snuke has a very long calendar. It has a grid with $n$ rows and $7$ columns. One day, he noticed that calendar has a following regularity.

The cell at the $i$-th row and $j$-th column contains the integer $7i+j-7$.

A good sub-grid is a $3 \times 3$ sub-grid and the sum of integers in this sub-grid mod $11$ is $k$.

How many good sub-grid are there? Write a program and help him.

Input

The input is given from standard input in the following format.

$n \quad k$

Output

Print the number of good sub-grid. If there are no solution exists, print 0.

Constraints

$1 \le n \le 10^9$

$0 \le k \le 10$

Subtasks

Subtask 1 [ $150$ points ]

The testcase in the subtask satisfies $1 \le n \le 100$.

Subtask 2 [ $100$ points ]

There are no additional constraints.

Sample Input 1

7 7

Sample Output 1

2

In this case, the calendar likes this matrix.

Sun.

Mon.

Tue.

Wed.

Thu.

Fri.

Sat.

Week 1

1

2

3

4

5

6

7

Week 2

8

9

10

11

12

13

14

Week 3

15

16

17

18

19

20

21

Week 4

22

23

24

25

26

27

28

Week 5

29

30

31

32

33

34

35

Week 6

36

37

38

39

40

41

42

Week 7

43

44

45

46

47

48

49

The cell at $i$-th row and $j$-th column is denoted $(i, j)$.

If upper-left is $(1, 5)$, the sum of integers is $5+6+7+12+13+14+19+20+21=117$.

If upper-left is $(3, 2)$, the sum of integers is $16+17+18+23+24+25+30+31+32=216$.

Therefore, there are 2 good sub-grids.

Sample Input 2

6 0

Sample Output 2

2

If upper-left is $(1, 3)$ or $(4, 4)$, it is a good sub-grid.

Sample Input 3

18 10

Solution in C++ (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
int n,f,s[14][8],t,h,ans;
int main()
{
    scanf("%d%d",&n,&f);
    for(int i=1;i<=11;i++)
    for(int j=1;j<=7;j++)
    {
        t++;
        s[i][j]=t;
    }
    for(int i=1;i<=11;i++)
        for(int j=1;j<=7-3+1;j++)
        {
            t=0;
            for(int k=0;k<3;k++)
            for(int l=0;l<3;l++)
                t+=s[(i+k-1)%11+1][j+l];
            if(t%11==f)
			    h++;
        }
    ans+=max(0,(n/11-1))*h;
    if(n/11>0)
	    n-=(n/11-1)*11;
    for(int i=1;i<=n-3+1;i++)
        for(int j=1;j<=7-3+1;j++)
        {
            t=0;
            for(int k=0;k<3;k++)
                for(int l=0;l<3;l++)
                    t+=s[(i+k-1)%11+1][j+l];
            if(t%11==f)ans++;
    }
    printf("%d\n",ans);
    return 0;
}