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())


As = []
xys = []
for i in range(N):
    tmp = []
    As.append(int(input()))
    for i in range(As[-1]):
        tmp.append(list(map(int, input().split())))
    
    xys.append(tmp)

#bit演算の準備
bit_list = list(itertools.product(range(2), repeat = N))

#loop
ans = 0 #ansの初期設定
for bit in bit_list:
    unkind_count = 0
    for Aindex, A in enumerate(As):
        if(bit[Aindex] == 0):
            continue #Unkindの意見は参考にしない
        for xy in xys[Aindex]:
#            print(xy)
            #真偽値を値化
            honest_to_unkind = (xy[1] == 0 and bit[xy[0]-1] == 1)
            unkind_to_honest = (xy[1] == 1 and bit[xy[0]-1] == 0)

            if(honest_to_unkind or unkind_to_honest):
                unkind_count +=1
    
        if(unkind_count == 0):
#            print("yatta")
            ans = max(ans, sum(bit))

print(ans)