Let circle $O$ have radius $5$ with diameter $\overline{AE}$. Point $F$ is outside circle $O$ such that lines $\overline{FA}$ and $\overline{FE}$ intersect circle $O$ at points $B$ and $D$, respectively. If $FA = 10$ and $m \angle FAE = 30^\circ$, then the perimeter of quadrilateral $ABDE$ can be expressed as $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$, where $a, b, c$, and $d$ are rational. Find $a + b + c + d$.

Let's think step by step and output the final answer within \boxed{}.
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