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 (3.8.2) (judge verdict: Wrong Answer)

n = int(input())

a_list = []
person_list = []
syogen_list = []
ans = 0

for i in range(0, n):
    w = []
    v = []
    a = int(input())
    for j in range(0, a):
        x,y = map(int,input().split())
        w.append(x)
        v.append(y)
    person_list.append(w)
    syogen_list.append(v)
#print(person_list)
#print(syogen_list)

for bit in range(1 << n):
    break_flag = 0
    bit_1_cnt = 0
    for i in range(n):
        #print(bin(bit))
        if break_flag:
            break
        if (bit >> i) & 1: #人iが正直
            for j in range(len(syogen_list[i])):
                if syogen_list[i][j] != ((bit>>(person_list[i][j]-1))&1): #正直としているのに証言が食い違えばbreak
                    break_flag = 1
                    #print("1_break",bit>>(person_list[i][j]-1))
                    break
                if j == len(syogen_list[i])-1:
                    bit_1_cnt += 1
#        else: #人iが嘘つき
#            for j in range(len(syogen_list[i])):
#                if syogen_list[i][j] == ((bit>>(person_list[i][j]-1))&1):
#                    break_flag = 1
#                    #print("0_break")
#                    break
        if break_flag==0 and i==n-1:
            ans = max(ans, bit_1_cnt)
            #print(bin(bit), bit_1_cnt)
            #print("correct")
print(ans)