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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program HonestOrUnkind2
  implicit none
  integer xy(15,15),a(15)
  integer n,ans,i,j,x,y,bit
  logical ok
  read(*,*) n
  xy(:,:)=-1
  do i=1,n
     read(*,*) a(i)
     do j=1,a(i)
        read(*,*) x,y
        xy(i,x)=y
     end do
  end do
  do bit=1,ishft(1,n)-1
     ok=.true.
     iloop:do i=1,n
        if(iand(bit,ishft(1,i-1))==0) cycle
        do j=1,a(i)
           if((xy(i,j)/=-1).and.(iand(bit,ishft(i,j-1))/=xy(i,j))) then
              ok=.false.
              exit iloop
           end if
        end do
     end do iloop
     if(ok) then
        ans=max(ans,popcnt(bit))
     end if
  end do
  print*,ans
end program HonestOrUnkind2