Assume $p=\frac{1+\theta}{n}$ where $\theta\gg n^{-\frac{1}{3}}$. Prove that a.a.s. the random graph $G_{n,p}$ almost surely has exactly one complex component. (A component $C$ is complex if it contains at least two distinct cycles. In terms of edges, $C$ is complex iff it contains at last $|C|+1$ edges).
This is problem 2.4.11 from Introduction to Random Graphs by Frieze and Karonski (phrased in terms of $G(n,p)$ rather than $G(n,m)$ for convenience).
My idea is to use staged exposure: write $G_{n,p}=G_{n,p_1}\cup G_{n,p_2}$ where $p_1=\frac{1+\theta/2}{n}$ and $p_2\sim \frac{\theta}{2n}$. Intuitively, it should be that any complex component of $G_{n,p_1}$ is sufficiently large, so adding the edges of $G_{n,p_2}$ will connect them. (Also one should take care of showing that the new edges do not form any new complex components). $p_2\gg \frac{1}{n^{4/3}}$ so it seems like "sufficiently large" should mean "having $\Omega(n^{2/3})$ vertices" (since $A=O(n^{2/3})$ implies $(1-p_2)^{A^2}\rightarrow 0$).
I tried working out a solution with this idea, but things didn't quite add up. I'd appreciate your help with a hint to further direct me towards the right outline of the proof.