acid base - Which molecule is more acidic: CHF3 or CHBr3?

I know that fluorine is more electronegative than bromine. However, because of the size of bromine, it is more stable with a negative charge. In the case of $\ce{HF}$ vs. $\ce{HBr}$, to me, $\ce{HBr}$ is without a doubt the stronger acid. When it comes to comparing $\ce{CHF3}$ with $\ce{CHBr3}$, we are supposed to compare the stability of their conjugate bases.

In $\ce{CF3^-}$, the fluorine can hold the electronegative charge quite well. It also exerts a strong inductive effect.

In $\ce{CBr3^-}$, the bromine is less electronegative than the fluorine but is more polarizable, so I think it can hold the negative charge more efficiently.

With this understanding, I am concluding that $\ce{CBr3^-}$ is more stable which would mean $\ce{CHBr3}$ is the stronger acid. However, I've been told otherwise.

I'm aware that the $\ce{C-Br}$ bond length is longer than the $\ce{C-F}$ bond length. I'm wondering if that short bond length causes $\ce{CF3^-}$ to be more stable than $\ce{CBr3^-}$. If this is true, is it okay for me to think about $\ce{CF3^-}$ dissociating vs. $\ce{CBr3^-}$ dissociating? In that scenario, I believe $\ce{CBr3^-}$ dissociates more easily than $\ce{CF3^-}$, meaning $\ce{CF3-}$ is more stable.

Note: This logic wouldn't apply to the instance of $\ce{HO-}$ vs. $\ce{HS-}$ because $\ce{HS-}$ is more stable than $\ce{HO-}$ (even though $\ce{H-O}$ bond length is shorter than $\ce{H-S}$ bond length).