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 (7.3.0) (judge verdict: Accepted)

import itertools
N = int(input())
xys=[]
for i in range(N):
    A = int(input())
    xy = []
    for i in range(A):
        xy.append(list(map(int,input().split())))
    xys.append(xy)
#print('xys:', xys)  ############


#正直者1,  不親切0
a = [0,1]
Ls = list(itertools.product(a, repeat=N))
#print('Ls:', Ls)  ###########

mujun=[0]*(N+1)
count=0
ans=0
for L in Ls:
    #print('L:        ', L)  ###########
    flag=0
    hito=1
    #count=0
    count=sum(list(L))
    mujun=[0] + list(L)
    for i in L:
        #print('i:', i)    ###########
        if i==1:
            for xy in xys[hito-1]:
                #print('xy:', xy)  #########
                if xy[1] == 1:
                    if mujun[xy[0]]==0:
                        count=0
                        #print('mujuuuuuunnnnnnn!')
                        flag=1
                        break
                    #else:
                        #count+=1
                        #mujun[xy[0]]=1
                else:
                    if mujun[xy[0]]==1:
                        count=0
                        #print('mujuuuuuuuuuuuuuuuuuuuun!')   ###########
                        flag=1
                        break
                    #else:
                        #mujun[xy[0]]=-1


                #print('count:', count)  ###########
                #print('mujun:', mujun)  ###########
        hito+=1
    if flag==0:
        ans=max(ans,count)
    #print('ans:', ans)  ###########

#print('ans:', ans) ###########
print(ans)