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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import itertools
n=int(input())
T,L=[[] for _ in range(n)],[[] for _ in range(n)]
for i in range(n):
  m=int(input())
  for j in range(m):
    x,y=map(int,input().split())
    if y==1:
      T[i].append(x-1)
    else:
      L[i].append(x-1)
ans=0

for num in range(1,n+1):
  for honest in itertools.combinations(range(n),num):
    honest=list(honest)
    nothonest=[i for i in range(n) if i not in honest]
    t1=list(itertools.chain.from_iterable([T[i] for i in honest]))
    l1=list(itertools.chain.from_iterable([L[i] for i in honest]))
    t2=list(itertools.chain.from_iterable([L[i] for i in nothonest]))
    l2=list(itertools.chain.from_iterable([T[i] for i in nothonest]))
    t=set(t1)&set(t2)
    l=set(l1)&set(l2)
    h=all(i in t for i in honest) and all(i not in l for i in honest)
    if h==True:
      ans=max(ans,num)

print(ans)