Sentence


Parse

How
(S/S)/(S[adj]\NP)
 
long
S[adj]\NP
 
S/S
> 0
have
(S[dcl]/(S[pt]\NP))/NP
 
you
NP
 
S[dcl]/(S[pt]\NP)
> 0
been
(S[pt]\NP)/PP
 
in
PP/NP
 
Kobe
N
 
NP
*
?
.
 
NP
.
PP
> 0
S[pt]\NP
> 0
S[dcl]
> 0
S[dcl]
> 0

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