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First note that it is not necessary for $B$ to guess the correct $n$. To see this, you can think of the PPT adversary $A'$ that on input $(y, 1^{n-1})$ outputs $A(y, 1^n)$. The sole purpose of the input $1^n$ is to allow $A$ to run in polytime in $n$. However, if the input (i.e., the output of $f$) to $B$ is of size, e.g., a constant $k$, then $B$ can not ...

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