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: Wrong Answer)

n = int(input())
ret = {}
for i in range(0, n):
    a = int(input())
    ret[i] = [[], []]
    for j in range(a):
        x, y = map(int, input().split())
        x -= 1
        if y == 1:
            ret[i][0].append(x)
        elif y == 0:
            ret[i][1].append(x)

dic = {}


def f(honest_p, honest_list):
    if honest_p in dic:
        return []
    dic[honest_p] = True
    # print('honest_list', honest_p, honest_list)
    if ret[honest_p][0] == []:
        return [honest_p]
    temp = [honest_p] + honest_list
    for i in honest_list:
        temp += f(i, ret[i][0])
    return temp


ans = 0
for i in range(0, n):
    dic = {}
    honest = list(set(f(i, ret[i][0])))
    # print('h', honest)
    dishonest = ret[i][1]
    for a in honest:
        for b in ret[a][1]:
            if b not in dishonest:
                dishonest.append(b)
    # print('d', dishonest)
    dishonest = list(set(dishonest))
    invalid = []
    for j in range(0, n):
        if i == j or j in honest or j in dishonest:
            continue
        val = True
        for k in ret[j][1]:
            if k in honest:
                val = False
        for k in ret[j][0]:
            if k in dishonest:
                val = False
        if not val:
            invalid.append(j)
    # print('invalid', invalid)
    import itertools
    lst = list(itertools.product([0, 1], repeat=n))
    for string in lst:
        string = list(string)
        for j in invalid:
            string[j] = 0
        # print(honest, dishonest, string)
        if list(set(honest) & set(dishonest)) == []:
            # print(string)
            ans = max(ans, string.count(1))
print(ans)