Let $f(x) = \frac{x-1}{x+1}$ and $f^{n}(x)$ denote the $n-\text{fold}$ composition of $f$ with itself. That is, $f^1(x) =f(x)$ and $f^n(x) = f(f^{n-1}(x))$. Find $f^{2007}(2)$.

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