2 int a
,b
,s
,t
;char c
[20000],*m
,*n
,*p
,*q
=c
;long long d
[6];void u(long long e
){
3 for(t
=0;e
/=2;t
++);}int main(void){for(q
+=fread(m
=n
=p
=c
,1,20000,stdin
);q
-m
;)
4 *n
-10?*p
-10?n
+=(++p
-m
)&1:(d
[s
]|=1ll<<((*m
&31)+26*(*m
<95)),d
[s
+1]|=1ll<<
5 ((*n
&31)+26*(*n
<95)),m
++,n
++):*p
==10?m
=n
=++p
,u(d
[s
]&d
[s
+1]),a
+=t
,
6 ((s
+=2)==6?u((d
[0]|d
[1])&(d
[2]|d
[3])&(d
[4]|d
[5])),b
+=t
,
7 s
=d
[0]=d
[1]=d
[2]=d
[3]=d
[4]=d
[5]=0,p
:p
):p
;printf("%d %d",a
,b
);}
