Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture:

<image>

Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap.

Input

The only line contain two integers m and d (1 ≤ m ≤ 12, 1 ≤ d ≤ 7) — the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday).

Output

Print single integer: the number of columns the table should have.

Examples

Input

1 7


Output

6


Input

1 1


Output

5


Input

11 6


Output

5

Note

The first example corresponds to the January 2017 shown on the picture in the statements.

In the second example 1-st January is Monday, so the whole month fits into 5 columns.

In the third example 1-st November is Saturday and 5 columns is enough.
month, first_day = (int(x) for x in input().split())

month_to_len = {1: 31, 2: 28,
                3: 31, 4: 30,
                5: 31, 6: 30,
                7: 31, 8: 31,
                9: 30, 10: 31,
                11: 30, 12: 31
               }
               
len_to_start = {28: 1, 
                30: 6, 
                31: 5
               }
               
len_to_min_columns = {28: 4,
                      30: 5,
                      31: 5
                     }

month_len = month_to_len[month]
columns_nr = len_to_min_columns[month_len]
if first_day > len_to_start[month_len]:
    columns_nr += 1
    print(columns_nr)
else:
    print(columns_nr)
    
