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)

def bin_(n):
    return str(bin(n))[2: ]

N = int(input())
syougen = [[] for i in range(N)]
for i in range(N):
    a = int(input())
    for j in range(a):
        x,y = input().split()
        x,y = int(x)-1,int(y)
        syougen[i].append((x,y))

start = 2**N-1
max_ = 0
for i in range(2**N):
    num = i
    uso = 0
    
    keta=bin(num)[2:]
    
    while(len(keta)<N):
        keta = "0"+keta
    print(keta)
    for j in range(N):
        if(keta[j]=="1"):
            for p in range(len(syougen[j])):
                
                x,y = syougen[j][p]
                if(keta[x] != str(y)):
                    uso = 1
                    break
    
    suuji = str(bin(num))[2:]
    count  = 0
    for i in range(len(suuji)):
        if(suuji[i]=="1"):
            count += 1
            
    if(uso==0):
        max_ = max(max_,count)
        
print(max_)