Conformal invariance in Toda field theories

A standard Toda field theory action will be of the shape:

$$S_{\text{TFT}} = \int d^2 x~ \Bigg( \frac{1}{2} \langle\partial_\mu \phi, \partial^\mu \phi \rangle - \frac{m^2}{\beta^2} \sum_{i=1}^r n_i e^{\beta \langle\alpha_i, \phi\rangle} \Bigg)$$

Where r is the rank of the algebra with simple roots $\alpha_i$. Adding the affine extension to the algebra corresponds to adding an extra root $\alpha_0$, and I read everywhere (e.g. Mussardo's stat. field theory book, and this PhD thesis: https://arxiv.org/abs/hep-th/0008200) that while normal Toda field theories are conformally invariant, this added root destroys the invariance.

My first question is: How is the non-affine case a CFT? When expanding the exponential to look at the quadratic terms, we get a mass-squared operator $M^2_{ij}= m^2 \sum_{k=1}^r n_i (\alpha_k)^i (\alpha_k)^j$. I don't see why this apparently is always zero in the non-affine case.

Second question: How can I see that the extra root finally makes these masses nonzero.

Thanks!

In conformal Toda theory all the interaction terms have dimension $1$ and are therefore marginal. This does not yet show that the theory is conformal: a sum of marginal operators is not always exactly marginal. Intuitively, the idea here is that in a theory of $b$ bosons, a sum of $r$ independent marginal exponential interaction terms is still marginal if $r\leq b$. In conformal Toda theory you have $r=b$ so you do have conformal symmetry. In the affine case you have $r=b+1$ so you lose exact conformal symmetry.
The bound on the number of interaction terms comes from trying to perturbatively compute corrections to correlation functions. You are perturbing a free bosonic theory where momentum conservation constrains the contributions of interaction terms. There are $b$ momentum conservation rules, which completely determine the contributions of all interaction terms to a given correlator if $r\leq b$. This prevents conformal symmetry from being broken.