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)

def counter(x):
    if x == 0:
        return 0
    return counter(x >> 1) + (x & 1)


N = int(input())
A = [0 for i in range(20)]
x = [[0 for i in range(20)] for j in range(20)]
y = [[0 for i in range(20)] for j in range(20)]

for i in range(1, N + 1):
    A[i] = int(input())
    for j in range(1, A[i] + 1):
        x[i][j], y[i][j] = list(map(int, input().split()))

# 2^N 通りをbit全探索
# kが正直者 = 2進数表記のk-1桁目が1 (1 <= k <= N)
ans = 0
for bits in range(1, 1 << N):
    ok = True
    for i in range(1, N + 1):
        # 不親切な人の証言は無視
        if ((bits >> (i - 1)) & 1) == 0:
            continue
        for j in range(1, A[i] + 1):
            # x_ij が正直者であることが偽であれば False
            if ((bits >> (x[i][j] - 1)) & 1) != y[i][j]:
                ok = False
    if ok:
        ans = max(ans, counter(bits))

print(ans)