Symmetric extension to Poisson Mass Function - Conditional Variance

Let $I=\pm1$ with equal chance. Let $Y$ be independent of I and have distribution $\textrm{Po}(\lambda)$. Consider $\displaystyle X\stackrel{d}{=}IY$. Use the conditional variance formula to compute $\textrm{Var} X$.

My attempt: $\textrm{Var} X=\mathbb{E}(\textrm{Var}(X|I))+\textrm{Var}(\mathbb{E}(X|I))$

For $X|I$, given $I=-1$, $X$ is equal in distribution to $-Y$. Given $I=1$, $X$ is equal in distribution to $Y$. So $X|I = \mathrm{sgn}(I)Y = IY$. But I'm stuck here, because it gets me back to where I started. Can someone help me? Is what I've done right so far?

If you condition on $I$ then you can treat $I$ as a constant when computing the conditional expectation/variance. So $$E[X \mid I] = E[IY \mid I] = I E[Y \mid I] = IE[Y] = \lambda I,$$ and then you can compute $\text{Var}(\lambda I)$. Similarly, $$\text{Var}(X \mid I) = \text{Var}(IY \mid I) = I^2 \text{Var}(Y \mid I) = \text{Var}(Y) = \lambda.$$