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)

#ABC-147
#参考 https://atcoder.jp/contests/abc147/submissions/8896111
N = int(input())
evidences = [[] for i in range(N)]

for i in range(N):
    A = int(input())
    for _ in range(A):
        x, y = map(int, input().split())
        evidences[i].append((x-1, y))

result = 0
for i in range(1, 2**N):
    consistent = True
    for j in range(N):
        #すでに矛盾するパターンのiとわかっているならば飛ばす。
        if consistent == False:
            break
        #jが正しいと仮定する。(もし0ならcontinueで飛ばされる。)
        if (i >> j) & 1 == 0:
            continue
        #jの証言を一個ずつ確認していく。
        for x, y in evidences[j]:            
            if (i >> x) & 1 != y:
                consistent = False
                break
    #全ての人の証言で矛盾がないパターンのiだったら、1の数(=正直者の数)をresultに記録しておく。
    if consistent == True:
        result = max(result, bin(i)[2:].count("1"))

print(result)