In $\triangle A B C$, $\overline { A B } = \overline { B C } = 3$ and $\cos \angle A B C = - \frac { 1 } { 8 }$. On the circumcircle of $\triangle A B C$ there is a point $D$ satisfying $\overline { B D } = 4$ and $\overline { A D } \leq \overline { C D }$. Then $\overline { C D } = $ (17-1) $+$ $\sqrt{\text{(17-2)}}$. (Express as a simplified radical form.)