Given a positive integer $A$ written in decimal expansion: $(a_{n},a_{n-1}, \ldots, a_{0})$ and let $f(A)$ denote $\sum^{n}_{k=0} 2^{n-k}\cdot a_k$. Define $A_1=f(A), A_2=f(A_1)$. Prove that:

[b]I.[/b] There exists positive integer $k$ for which $A_{k+1}=A_k$.
[b]II.[/b] Find such $A_k$ for $19^{86}.$