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 sys

def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))  #空白あり

N = I()
A = []
xy = [[] for i in range(N)]
for i in range(N):
    A.append(I())
    for j in range(A[i]):
        x,y = LI()
        xy[i].append([x-1,y])

# 人は0-index
# 各人が正直者か不親切な人かで全探索、1…正直者、0…不親切な人

ans = 0
for i in range(2**N):
    B = [[] for _ in range(N)]
    a = 0  # 正直者の人数
    for j in range(N):
        if (i>>j) & 1:  # jが正直者
            B[j].append(1)
            a += 1
            for k in range(A[j]):
                B[xy[j][k][0]].append(xy[j][k][1])
        else:  # jが不親切な人
            B[j].append(0)
    if sum(len(set(B[j])) for j in range(N)) != N:  # この場合矛盾
        continue
    else:
        ans = max(ans,a)

print(ans)