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 checker(bbit,listN,N):
    i = 0
    j = 0
    ng = False
    #print(N)
    #print(listN)
    while N > i :
        j=0
        checkFig = 3
        ibit = 1 << i
        while N > j :
            cbit = 1 << j
            if bbit & cbit == 0 :
                #print("A:{} {}".format(j,i))
                pass
            elif listN[j][i] == 3 :
                #print("B:{} {}".format(j,i))
                pass
            elif (listN[j][i] == 0) and (bbit & ibit != 0):
                ng = True
                break
            elif checkFig == 3 :
                #print("C:{} {}".format(j,i))
                checkFig = listN[j][i]
            elif checkFig != listN[j][i] :
                ng = True
                #print("D:{} {}".format(j,i))
                break
            j += 1
        if ng :
            #print(ng)
            break
        else :
            i +=1
    return (not ng)


def main():
    N = int(input())
    listA = [[3 for j in range(N)] for k in range(N)]
    n = 0
    while N > n :
        a = int(input())
        i = 0
        while a > i:
           x,y  = map(int,input().split())
           listA[n][x-1] = y
           i += 1
        n += 1
        
    honestBit = pow(2, N)-1
    honest = N
    while honest > 0:
        ok = checker(honestBit,listA,N)
        if ok :
            break
        honestBit -= 1
        honest -= 1

    print(honest)

main()