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)

n = int(input())
l = []
A = []
for i in range(n):
    a = int(input())
    A += [a]
    l_1 = []
    for j in range(a):
        l_1 += [list(map(int, input().split()))]
    l += [l_1]

ans = 0

for i in range(1 << n):
    honest = []
    liar = []
    for j in range(n):
        if ((i >> j) & 1):
            if j + 1 in liar:
                break
            elif j + 1 not in honest:
                honest += [j + 1]
            sugg = l[j]

            for k in sugg:
                if k[1] == 1:
                    if k[0] in liar:
                        break
                    elif k[0] not in honest:
                        honest += [k[0]]
                else:
                    if k[0] in honest:
                        break
                    elif k[0] not in liar:
                        liar += [k[0]]
            else:
                continue
            break

        else:
            if j + 1 in honest:
                break
            elif j + 1 not in liar:
                liar += [j + 1]
    else:
        if ans < len(honest):
            ans = len(honest)

print(ans)