Let $$ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b \quad \text{and} \quad H: \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1. $$ Let the distance between the foci of $E$ and the foci of $H$ be $2\sqrt{3}$. If $a - A = 2$, and the ratio of the eccentricities of $E$ and $H$ is $\frac{1}{3}$, then the sum of the lengths of their latus rectums is equal to: