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.
n=input()
n=n.split()
b=0
if int(n[0])==1 or int(n[0])==3 or int(n[0])==5 or int(n[0])==7 or int(n[0])==8 or int(n[0])==10 or int(n[0])==12:
    a=31
    a=a-(8-int(n[1]))
    if a%7==0:
        b=1+(a//7)
        print (b)
    else:
        b=1+((a//7)+1)
        print(b)
elif int(n[0])==4 or int(n[0])==6 or int(n[0])==9 or int(n[0])==11:
    a=30
    a=a-(8-int(n[1]))
    if a%7==0:
        b=1+(a//7)
        print (b)
    else:
        b=1+((a//7)+1)
        print(b)
else:
    a=28
    a=a-(8-int(n[1]))
    if a%7==0:
        b=1+(a//7)
        print (b)
    else:
        b=1+((a//7)+1)
        print(b)
