def n_highest_grossing_actors(actors, n):
    """
    Query helper to get the the n highest grossing ACTORS
    :param actors: Dictionary (name of actor --> Actor node)
    :param n: Number of highest grossing ACTORS to find
    :return: List of n highest grossing ACTORS, sorted highest to lowest. If n > length of ACTORS,
    all the ACTORS are returned ordered highest to lowest in terms of grossing value.
    """
    actors_sorted_by_gross_value = sorted(actors, key=lambda x:actors[x].get_grossing_value(), reverse=True)
    return actors_sorted_by_gross_value[:min(n, len(actors_sorted_by_gross_value))]