int pre
[MX
],diao
[MX
];
void solve()
{
int n
;cin
>>n
;
rpp(i
,n
)
{
int m
;cin
>>m
;
cin
>>pre
[i
];
rpp(j
,m
-1)
{
int x
;cin
>>x
;
if(x
>0&&(x
!=pre
[i
]))
diao
[i
]=1,pre
[i
] = x
;
if(x
<0)
pre
[i
]+=x
;
}
}
int sum
= 0,luo
= 0,lian
= 0;
rpp(i
,n
)
{
sum
+=pre
[i
];
if(diao
[i
]) ++luo
;
int l
= i
-1,r
=i
+1;
if(i
==1) l
=n
;
if(i
==n
) r
=1;
if(diao
[i
]&&diao
[l
]&&diao
[r
]) ++lian
;
}
cout
<<sum
<<" "<<luo
<<" "<<lian
<<endl
;
}
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