Add solution for 3e
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$\unit[655]{bit}$ input to map to 101 zeros. Therefore, the whole search
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space is $2^{101}$ and the probability with one query is $1/2^{101}$.
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\item \TODO
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\item If the adversary makes one inverse query, the probability that it hits
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a preimage is $1/2^{55+90} = 1/2^{145}$. This is due to the fact that the
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first $\unit[55]{bits}$ and the last $\unit[90]{bits}$ have to be zeros.
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\item \TODO
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