If $\sum\limits_{r=1}^{13}\left\{\frac{1}{\sin \left(\frac{\pi}{4}+(r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{r \pi}{6}\right)}\right\}=a \sqrt{3}+b, a, b \in Z$, then $a^2+b^2$ is equal to :