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.
def check_month(day, month):
    return sum(months[:month]) <= day < sum(months[:month + 1])

A = [0] * 365
months = [31, 	28, 	31, 	30, 	31, 	30, 	31, 	31, 	30, 	31, 	30, 	31]
for i in range(365):
    A[i] = i % 7
m, d = map(int, input().split())
m -= 1
first_day = 0
while not check_month(first_day, m):
    first_day += 1
diff = d - A[first_day]
for i in range(365):
    A[i] += diff
    A[i] %= 7
sundays = 0
for i in range(365):
    if check_month(i, m):
        if A[i] == 0:
            sundays += 1
while check_month(first_day, m):
    first_day += 1
first_day -= 1
if A[first_day] != 0:
    sundays += 1
print(sundays)