Capacity of a 2-user discrete memoryless channel

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I am trying to understand the capacity of the following 2 user multiple access channel with input and output alphabets $\mathcal{X}_{1}=\mathcal{X}_{2}=\mathcal{Y}=\{0,1\}$.

The channel transition probabilities are as below:

If $\left(X_{1}+ X_{2}\right)=$ $0$, then $Y=0$ with probability $p_1$.

If $\left(X_{1} + X_{2}\right)=1$, then $Y=0$ with probability $p_2$.

If $\left(X_{1} + X_{2}\right)=2$, then $Y=0$ with probability $p_3$.

I want to compute the capacity of this channel?

My attempt: I know that the capacity of a multiple-access channel $\left(\mathcal{X}_{1} \times \mathcal{X}_{2}, p\left(y \mid x_{1}, x_{2}\right), \mathcal{Y}\right)$ is the closure of the convex hull of all $\left(R_{1}, R_{2}\right)$ satisfying $$ \begin{aligned} R_{1} &<I\left(X_{1} ; Y \mid X_{2}\right) \\ R_{2} &<I\left(X_{2} ; Y \mid X_{1}\right) \\ R_{1}+R_{2} &<I\left(X_{1}, X_{2} ; Y\right) \end{aligned} $$

By symmetry I can assume, $Pr(X_1=0)=Pr(X_2=0)=q$ achieves capacity and now I need to find $q$.

Can someone help me proceed further? I tried to expand the mutual information terms, but I was not able to proceed further.

It would be great if someone can characterize the capacity region and the capacity achieving input distribution.