In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $\overline{AB}$ at $L$. The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$. If $LI=2$ and $LD=3$, then $IC= \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.