Sentence


Parse

Thanks
N/PP
 
for
PP/(S[ng]\NP)
 
being
(S[ng]\NP)/(S[adj]\NP)
 
here
S[adj]\NP
 
.
.
 
S[adj]\NP
.
S[ng]\NP
> 0
PP
> 0
N
> 0
NP
*

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