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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