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)

#標準入力受け取り
N = int(input())
shougen = []
A_lis = []
for _ in range(N):
    A = int(input())
    A_lis.append(A)
    minishougen = []
    for j in range(A):
        xy = list(map(int, input().split()))
        minishougen.append(xy)
    shougen.append(minishougen)

#ビットの配列作る
bit_lis = []
for i in range(2 ** N):
    mini_bit_lis = []
    for j in range(N):
        mini_bit_lis.append(i % 2)
        i = i // 2
    bit_lis.append(mini_bit_lis)
    
#ビットの配列について全探索を行う
ans = 0
for i in range(2 ** N):
    flag = 0
    for l in range(N):
        if bit_lis[i][l] == 1:
            for k in range(A_lis[l]): 
                if shougen[l][k][1] != bit_lis[i][shougen[l][k][0] - 1]:
                    flag = 1
                    break
        if flag == 1:
            break
    if flag == 0:
        tmp = sum(bit_lis[i])
        if tmp > ans :
            ans = tmp
print(ans)