In triangle $ABC,$ $D$ is on $\overline{AB}$ and $E$ is on $\overline{BC}$ such that $AD = 3,$ $BD = 2,$ $BE = 3,$ and $CE = 4.$  Let $F$ be the intersection of $\overline{AE}$ and $\overline{CD}.$  Find $\frac{EF}{FA}.$The answer is in the form rac{m}{n}, where gcd(m, n) = 1. Please provide the value of m + n.

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