For all positive integers \(n\), the totient function \(\phi (n)\) denotes the number of positive integers \(\leq n\) coprime to \(n\).

It turns out that if we apply the totient function successively on any positive integer \(n>1\), after finitely many operations we get the number \(1\). In other words, for all positive integers \(n>1\), there exists a positive integer \(m\) such that \(\underbrace{\phi ( \phi ( \phi ( \cdots \phi (n ) \cdots )))}_{m \text{ times}}= 1\).

For all \(n \in \mathbb{N}\), let \(f(n)\) denote the minimum number of times we need to apply the totient function successively on \(n\) to get the number \(1\). Find the last three digits of \(\displaystyle \sum \limits_{n=1}^{2014} f(n)\).

**Details and assumptions**

As an explicit example, here's how you'd find \(f(5)\): \[\begin{array}{rcl} \phi (5)& = & 4 \\ \phi (\phi (5)) & = & 2 \\ \phi (\phi (\phi (5))) & = & 1 \\ \end{array} \] Note that we had to apply the totient function three times on \(5\) to get the number \(1\), so \(f(5)=3\).

By convention, \(f(1)=0\).

Just to clarify, this is a computer science problem.

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