An unbiased coin is tossed repeatedly until two consecutive heads are obtained. Suppose these occur on the $(M-1)$th and $M$th toss.
Let $P(n)$ be the probability that $M$ is divisible by $n$. For example, the outcomes HH, HTHH, and THTTHH all count towards $P(2)$, but THH and HTTHH do not.

You are given that $P(2)=\frac{3}{5}$ and $P(3)=\frac{9}{31}$. Indeed, it can be shown that $P(n)$ is always a rational number.

For a prime $p$ and a fully reduced fraction $\frac{a}{b}$, define $Q(\frac{a}{b},p)$ to be the smallest positive $q$ for which $a\equiv bq(\mathrm{mod}p)$.
For example $Q(P(2),109)=Q(\frac{3}{5},109)=66$, because $5\cdot 66=330\equiv 3(\mathrm{mod}109)$ and $66$ is the smallest positive such number.
Similarly $Q(P(3),109)=46$.

Find $Q(P({10}^{18}),1000000009)$.