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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
typedef long long ll;

ll N,K;

int sum[12][11];

void solve(){
	ll a,b;
	if (N<=2){
		printf("0\n");
		return;
	}
	a=N-2;
	b=a%11;
	ll res=0;
	res=sum[11][K]*(a/11);
	REP(i,b){
		res+=sum[i][K];
	}
	printf("%lld\n",res);
}

int main(){
	int F=13;
	FOR(i,2,F){
		FOR(j,2,7){
//			printf("%3d ",(7*i+j-7)*9%11);
			sum[i-2][(7*i+j-7)*9%11]++;
		}
//		printf("\n");
	}
	REP(i,11){
		REP(j,11){
			sum[11][i]+=sum[j][i];
		}
	}
	scanf("%lld %lld",&N,&K);
	solve();
	/*
	FOR(i,2,18){
		FOR(j,2,7){
			fprintf(stderr,"%3d ",(7*i+j-7)*9%11);
		}
		fprintf(stderr,"\n");
	}
	*/
	return 0;
}