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;
using Int = long long;
int main() {
    cin.tie(nullptr); ios::sync_with_stdio(false);
    int N, K; cin >> N >> K;
    Int ans[11] = {};
    Int s = (1 + 2 + 3 + 8 + 9 + 10 + 15 + 16 + 17) % 11;
    for (int i = 0; i < 5; i++) {
        Int cnt = max(0, N - 2);
        for (int j = 0; j < 11; j++) {
            ans[j] += cnt / 11;
        }
        Int t = s;
        for (int j = 0; j < cnt % 11; j++) {
            ans[t] += 1;
            t = (t + 3 * 3 * 7) % 11;
        }
        s = (s + 3 * 3) % 11;
    }
    cout << ans[K] << endl;
    return 0;
}