Finding the UMVUE for $p^2$ when $X_1,...,X_n \sim Ber(p)$

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I had a question regarding MLEs and UMVUEs.

My understanding of UMVUEs is the following.

Let $\hat{\theta}$ be the MLE of a distribution and $\hat{\theta}^*$ be the adjusted statistic based on the MLE that is now unbiased.

For $\hat{\theta}^*$ to be a UMVUE the distribution just has to be complete.

Using this idea, I formed the following hypothesis.

To find the UMVUE for $p^2$ when $X_1,...,X_n \sim Ber(p)$, I know that $\bar{X}$ is the MLE for $p$ and also unbiased.

$Ber(p)$ is a complete distribution, so according to the invariance of MLEs I am thinking that the estimator for $p^2$ should be $\bar{X}^2$ and that will become the UMVUE.

However, based on Lehmann-Sheffe's Theorem,

$$E[X_1X_2|\sum_{i=1}^nX_i=x] = \frac{\bar{X}(n\bar{X}-1)}{n-1}$$ should be the UMVUE.

I do not know why my solution is not coincident with the result of the theorem . . .

Thank you for your help.