Energy conversion efficiency of the steam engine.

The path followed by water in steam engines can be described as the following:

From A to B: Liquid water is compressed in pump, it receives a work $W_{A \rightarrow B}$.

From B to C: Water is heated in a boiler where it receives heat $Q$ without any work, and from where it gets out as vapor.

From C to D: In the turbine, water gives the amount of work $W_{C \rightarrow D}$ without gaining or losing heat.

From D to A: Water is cooled in a condenser, without work transfer, where it goes back to its original state and properties, to go the pump again to get compressed and so on...

Q: What is the energy conversion efficiency $\eta$ of the steam engine?

We know that it is the ratio of the useful output energy and the input energy.

In this case, the output energy is the work $W_{C \rightarrow D}$ done by the water during the phase $C \rightarrow D$ , and the input is the work of compression $W_{A \rightarrow B}$ and the heat Q.

In the solution of the problem, the energy conversion efficiency is expressed as: $$\eta = \frac{W_{A \rightarrow B} + W_{C \rightarrow D} }{Q}$$

Can you explain to me why it is not: $\eta = \frac{W_{C \rightarrow D} }{Q + W_{A \rightarrow B} }$ ?

Thank you.