for any (u,v,w) there exist only 1 vertex x such x in all 3 simple path's (u,v),(u,w),(v,w). basically you can proof that p(u,v)*p(u,w)*p(v,w)=n²*x where n natural. So (u,v,w) good if x = k². So for every such an x, you need to calc amount of that triples. That's 0 if x has one son, y1*y2 if x has 2 sons, (s(y)²-s(y²))/2+(s(y)³-3s(y²)s(y)+2s(y³))/6 if x has >=3 sons, where y is arr of number of descendants of the sons, s(yⁿ) = sum(yiⁿ) for every i.

why does number of G solvers decreases?...

the fact that I enjoyed C2 so much even while I failed to send it on time proofs it's very interesting problem anyway

I honestly wanna know it too. I don't though that pure interest is sexism or smthlth