How to show that the interaction $p + bar p rightarrow pi^0 + pi^0$ is forbidden in the electromagnetic and strong interaction?

Question

(I am at the introductory level of particle physics)
I know that the parity is conserved in strong and electromagnetic interaction, so I would like to show that the charge parity is violated.
I would like to show that the product of the charge conjugation parity of proton and its antiparticle is $-1$, which doesn't equal to the charge parity of $pi^0 + pi^0$
problem I have encountered:
by definition $hat C |p, Psirangle = C_p |p,Psirangle$ and $hat C |p, Psirangle = |bar p,Psirangle$ then $hat C^2 |p, Psirangle = C_p^2 |p,Psirangle = |p,Psirangle = C_bar p |bar p,Psirangle$
applying $bar C$ once again
$C_p|p,Psirangle = C_bar p |bar p,Psirangle implies C_p = C_bar p implies C_pC_bar p = 1$ which doesn't lead to the result I wanted.
What went wrong for me?

anna v · Answer

In this paper , at page 9:

4.1 Annihilation into two neutral mesons

Crystal Barrel has measured the branching ratios for antiproton proton  annihilation into two neutral light mesons from about $10^7$ annihilations into 0-prong (Amsler, 1993b). These data have been collected by vetoing charged particles with the PWC’s and the internal layers of the JDC.The lowestγ-multiplicity was four (e.g.π0π0,π0η) and the highest nine (e.g.ηω, with  η→3π0and ω→π0γ).

So the reaction is seen experimentally, table at page 54 . The possible diagrams discussed are also strong interaction ones, as discussed in paragraph 4.2, illustrated in fig. 9

How to show that the interaction $p + bar p rightarrow pi^0 + pi^0$ is forbidden in the electromagnetic and strong interaction?

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