Let unsigned long long total_cost=0.
For every edge E in tree total_cost +=cost_of_all_paths_lies_on_edge(E).
double ans=2.0*total_cost/n/(n-1) is overflow and and WA#19.
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Let double ans = 0.
For every edge E in tree ans+=2.0*cost_of_all_paths_lies_on_edge(E)/n/(n-1).
That is AC.
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Maybe if not multiply by two in the first scenario there is also AC,  I have not tested.