Let the function $ f(x) = \frac{x}{3} + \frac{3}{x} + 3 $, $ x \neq 0 $, be strictly increasing in $ (-\infty, \alpha_1) \cup (\alpha_2, \infty) $ and strictly decreasing in $ (\alpha_3, \alpha_4) \cup (\alpha_5, \alpha_s) $. Then $ \sum_{i=1}^{5} \alpha_i^2 $ is equal to: