Problem:
Score : 300 points

Problem Statement

There are N people numbered 1 to N. Each of them is either an honest person whose testimonies are always correct or an unkind person whose testimonies may be correct or not.

Person i gives A_i testimonies. The j-th testimony by Person i is represented by two integers x_{ij} and y_{ij}. If y_{ij} = 1, the testimony says Person x_{ij} is honest; if y_{ij} = 0, it says Person x_{ij} is unkind.

How many honest persons can be among those N people at most?

Constraints

All values in input are integers.

1 \leq N \leq 15

0 \leq A_i \leq N - 1

1 \leq x_{ij} \leq N

x_{ij} \neq i

x_{ij_1} \neq x_{ij_2} (j_1 \neq j_2)

y_{ij} = 0, 1

Input

Input is given from Standard Input in the following format:

N
A_1
x_{11} y_{11}
x_{12} y_{12}
:
x_{1A_1} y_{1A_1}
A_2
x_{21} y_{21}
x_{22} y_{22}
:
x_{2A_2} y_{2A_2}
:
A_N
x_{N1} y_{N1}
x_{N2} y_{N2}
:
x_{NA_N} y_{NA_N}

Output

Print the maximum possible number of honest persons among the N people.

Sample Input 1

3
1
2 1
1
1 1
1
2 0

Sample Output 1

2

If Person 1 and Person 2 are honest and Person 3 is unkind, we have two honest persons without inconsistencies, which is the maximum possible number of honest persons.

Sample Input 2

3
2
2 1
3 0
2
3 1
1 0
2
1 1
2 0

Sample Output 2

0

Assuming that one or more of them are honest immediately leads to a contradiction.

Sample Input 3

2
1
2 0
1
1 0

Sample Output 3

1

Solution in Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

type xy = record x,y:int64; 
                 end;
     p = record a:int64;
                t:array[1..20] of xy;
                end;
var i,j,l,maker,power,ans,n:int64;
    people:array[1..20] of p;
    check:array[1..20] of int64;
    flag :boolean;
    
function power2(q:int64):int64;
var po,k:int64;
begin
 po:=1;
 for k := 1 to q do po := po *2;
 power2 := po
end;

begin
read(n);
for i := 1 to n do
 begin
  read(people[i].a);
   for j := 1 to people[i].a do read(people[i].t[j].x,people[i].t[j].y)
 end;

power := power2(n);
for i := 0 to power-1 do
 begin
  maker := i;
  for j := 1 to n do
   begin 
    if odd(maker) then check[j] := 1
    else check[j]:=0;
    maker := maker >> 1
   end;
  for j := 1 to n do
   begin 
    flag := true;
    if (check[j] = 1) then
     begin
      for l := 1 to people[j].a do 
       if check[people[j].t[l].x] <> people[j].t[l].y then 
        begin
         flag := false;
         break
        end
     end
    else 
     begin
      for l := 1 to people[j].a do 
       if check[people[j].t[l].x] = people[j].t[l].y then 
        begin
         flag := false;
         break;
        end
     end;
    if not (flag) then break;
   end;
  
 if flag then break;
 end;

ans:=0;
if flag then
 begin
 for i := 1 to n do
  if check[i] = 1 then inc(ans)
 end
else ans := 0;

write(ans)

end.