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: Time Limit Exceeded)

n = int(input())
#ビット全探索を行う
import itertools
li = list(itertools.product([0, 1], repeat=n)) # ０が不親切もの、１が親切もの

#dd[i]:人iが人xをy（０なら不親切,１なら親切)だと思っていることを収納
from collections import deque, defaultdict
dd = defaultdict(lambda:deque())
for i in range(n):
    a = int(input())
    for j in range(a):
        x, y = map(int, input().split())
        dd[i].append((x-1, y))


from copy import deepcopy
max_sum = 0 #親切者の数の最大値
for i in li: # i には親切者のタプルが入っている
    sum_ = 0
    co = deepcopy(dd)
    vv = 0
    for k in range(len(i)): #全員の証言が矛盾しないかを考える
        if i[k] == 0: # 不親切ものの場合
            continue
        else: #以下k人目の証言を考える
            sum_ += 1
            length = len(co[k])
            c = 0
            while co[k]:
                x, y = co[k].popleft()
                if i[x] != y: #証言と一致していない場合
                    break
                else:
                    c+=1
            if c == length: #k人目の証言が全てあっている場合
                vv += 1
    if vv == sum_:
        max_sum = max(max_sum, vv)

print(max_sum)