
__magic_name__            =  {}





def 	SCREAMING_SNAKE_CASE__    (      __lowerCAmelCase    ,		__lowerCAmelCase    ,		__lowerCAmelCase							):

    # if we are absent twice, or late 3 consecutive days,
    # no further prize strings are possible
    if late == 3 or absent == 2:
        return 0

    # if we have no days left, and have not failed any other rules,
    # we have a prize string
    if days == 0:
        return 1

    # No easy solution, so now we need to do the recursive calculation

    # First, check if the combination is already in the cache, and
    # if yes, return the stored value from there since we already
    # know the number of possible prize strings from this point on
    snake_case__						=				(days, absent, late)
    if key in cache:
        return cache[key]

    # now we calculate the three possible ways that can unfold from
    # this point on, depending on our attendance today

    # 1) if we are late (but not absent), the "absent" counter stays as
    # it is, but the "late" counter increases by one
    snake_case__						=				_calculate(days - 1    ,		__lowerCAmelCase    ,		late + 1							)

    # 2) if we are absent, the "absent" counter increases by 1, and the
    # "late" counter resets to 0
    snake_case__						=				_calculate(days - 1    ,		absent + 1    ,		0							)

    # 3) if we are on time, this resets the "late" counter and keeps the
    # absent counter
    snake_case__						=				_calculate(days - 1    ,		__lowerCAmelCase    ,		0							)

    snake_case__						=				state_late + state_absent + state_ontime

    snake_case__						=				prizestrings
    return prizestrings





def 	SCREAMING_SNAKE_CASE__    (      __lowerCAmelCase = 30							):
    return _calculate(__lowerCAmelCase    ,		absent=0    ,		late=0							)


if __name__ == "__main__":
     print(solution())





from __future__ import annotations





def 	SCREAMING_SNAKE_CASE__    (      __lowerCAmelCase    ,		__lowerCAmelCase							):
    snake_case__						,					snake_case__						=				set(__lowerCAmelCase							), [start]

    while stack:
        snake_case__						=				stack.pop()
        explored.add(__lowerCAmelCase							)
        # Differences from BFS:
        # 1) pop last element instead of first one
        # 2) add adjacent elements to stack without exploring them
        for adj in reversed(graph[v]							):
            if adj not in explored:
                stack.append(__lowerCAmelCase							)
    return explored


__magic_name__            =  {
    '''A''': ['''B''', '''C''', '''D'''],
    '''B''': ['''A''', '''D''', '''E'''],
    '''C''': ['''A''', '''F'''],
    '''D''': ['''B''', '''D'''],
    '''E''': ['''B''', '''F'''],
    '''F''': ['''C''', '''E''', '''G'''],
    '''G''': ['''F'''],
}

if __name__ == "__main__":
     import doctest

     doctest.testmod()
     print(depth_first_search(G, '''A'''))



