Sentence


Parse

Can
(S[q]/(S[b]\NP))/NP
 
I
NP
 
S[q]/(S[b]\NP)
> 0
send
((S[b]\NP)/PP)/NP
 
a
NP/N
 
fax
N
 
NP
> 0
(S[b]\NP)/PP
> 0
from
PP/(S[adj]\NP)
 
here
S[adj]\NP
 
?
.
 
S[adj]\NP
.
PP
> 0
S[b]\NP
> 0
S[q]
> 0

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