Complex integral and Laurent series

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Could you help with solving this complex integral: $$\int_C z^3\exp{\left(\dfrac{-1}{z^2}\right)} dz$$ where $C$ is $|z|=5$. I am expecting that the Residue Theorem will be needed. The answer should be $i\pi$.

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Hint:

$$z^3e^{-1/z^2} = z^3 - \frac{1}{1!} z + \frac{1}{2!} \frac{1}{z} - \frac{1}{3!} \frac{1}{z^3} + \frac{1}{4!} \frac{1}{z^5} - \cdots$$

What happens when you take your contour integral of that sum? Which terms have non-zero residue?