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 Python (3.4.3) (judge verdict: Accepted)

n = int(input())
 
XY = []
for i in range(n):
    a = int(input())
    xy = []
    for j in range(a):
        xy.append(list(map(int, input().split())))
    
    XY.append(xy)
    
def check(i, xiyj, HoU):
    # 矛盾したらFalse 正直者と仮定した証言より 正直&正直者と仮定されている / 不親切&不親切と仮定されている True
    if ((i >> xiyj) & 1) ^ HoU:
        return False
 
    return True
    
max_honest = 0
for i in range(2 ** len(XY)):
    count = 0
    honest = [0] * len(XY)
    is_contradiction = True
    for j in range(len(XY)):
        # 正直者と仮定したときだけ採用
        if not ((i >> j) & 1): continue
        for k in range(len(XY[j])):
            xiyj = XY[j][k][0] - 1
            HoU = XY[j][k][1]
            
            if not check(i, xiyj, HoU):
                is_contradiction = False
 
    if is_contradiction:
        for j in range(len(XY)):
            if ((i >> j) & 1): count+=1 
                
        if count > max_honest:
            max_honest =count
            
print(max_honest)