Prim's algorithm on graph with weights of only $1$ and $2$ on each edge

Question

I have this version of Prim's algorithm

Prim$(G=(V,E),s\in V,w)\\ 1.\ d(s)\leftarrow 0;\forall u \neq s:d(u)\leftarrow \infty\quad \color{red}{O(|V|)}\\ 2.\ \forall u \in V:p(u)\leftarrow \text{null}\quad \color{red}{O(|V|)}\\ 3.\ s\leftarrow \emptyset,T\leftarrow \emptyset\quad \color{red}{O(1)}\\ 4.\ Q \leftarrow \text{Makeheap}(V)\quad \color{red}{O(|V|)}\\ 5.\ \text{while}\ Q\neq \emptyset\\ 6.\qquad u\leftarrow Q.\text{ExtractMin()}\quad \color{red}{O(\log(|V|))}\\ 7.\qquad S\leftarrow S\cup\{u\},\ T\leftarrow T\cup\{(u,p(u))\}\quad \color{red}{O(1)}\\ 8.\qquad \text{for each}\ (u,z)\in E,z\in Q\\ 9.\qquad\qquad \text{if}\ d(z)>w(u,z)\ \text{then}\\ 10.\qquad\qquad\qquad Q.\text{Remove}(z);Q.\text{Insert}(z,w(u,z))\quad \color{red}{O(\log(|V|))}\\ 11.\qquad\qquad\qquad d(z)\leftarrow w(u,z),\ p(z)\leftarrow \quad \color{red}{O(1)}\\ 12.\ \text{return }T$

Total time: $O(|E|\log(|V|))$

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Given a weighted, connected, simple undirected graph $G$ with weights of only $1$ and $2$ on each edge, why in this case the running time of the algorithm is $O(|E|+|V|\log(|V|))$?

I really not understand why the running time is not the same in this case, any help?