Sentence


Parse

I
NP
 
have
(S[dcl]\NP)/(S[to]\NP)
 
to
(S[to]\NP)/(S[b]\NP)
 
be
(S[b]\NP)/PP
 
on
PP/NP
 
my
NP/(N/PP)
 
way
N/PP
 
now
N\N
 
.
.
 
N\N
.
N/PP
< 1×
NP
> 0
PP
> 0
S[b]\NP
> 0
S[to]\NP
> 0
S[dcl]\NP
> 0
S[dcl]
< 0

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