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 PyPy3 (2.4.0) (judge verdict: Accepted)

N = int(input())
A = [None]*N
x = [[None]*N for _ in range(N)]
y = [[None]*N for _ in range(N)]

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

ans =0
for num in range(2**N):
    #bit = bin(num)#正直者リスト
    flag = 1 #これが1のままならOK
    
    #N人を順に見る
    for i in range(N):
        #下からi桁目を取り出す
        if (num>>i & 1)==1:#正直者なら検討
            
            #各発言に対して，NGケースを調査
            for j in range(N):
                #xijの対応に注意，文章では1から
                if x[i][j] != None and y[i][j]!=(num>>(x[i][j]-1) & 1):
                    flag = 0
        if flag ==0:
            break
    if flag ==1:
        ans = max(ans,bin(num).count("1"))
print(ans)