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.
from calendar import monthrange


def read(type=int):
    return type(input())


def read_arr(type=int):
    return [type(token) for token in input().split()]

def days(m):
    if m == 2:
        return 28
    if m in {1,3,5,7,8,10,12}:
        return 31
    return 30

def run():
    m, d = map(int, input().split())
    d -= 1
    w, n = monthrange(2017, m)
    ans = 1
    arr = [0 for i in range(7)]

    while n > 0:
        arr[d] = ans
        d += 1
        if d % 7 == 0:
            ans += 1
            d %= 7
        n -= 1
    print(max(arr))

def run2():
    m, d = read_arr()
    w, n = monthrange(2017, m)
    ans = 1
    i = d % 7
    for k in range(n):
        i = (i + 1) % 7
        if i % 7 == 1:
            ans += 1
    print(ans)

run()
