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 (3.4.3) (judge verdict: Accepted)

N = int(input())
# 出力結果[[(1, 1)], [(0, 1)], []]
evidences = [[] for _ in range(N)]
for i in range(N):
    # i人目の証言A個
    A = int(input())
    for _ in range(A):
        x, y = map(int, input().split())
        # 例1だと人1,2,3だからリストで0,1,2になる、yは0が0番目で問題ない
        evidences[i].append((x - 1, y))

result = 0
# 8パターンあって順に縦に回してく
for i in range(1, 2 ** N):
    consistent = True
    # 例題1では3桁を横に回す
    for j in range(N):
      # 例えば3個目の010をj=1ずらして0なら続ける1のフラグが立ってたら次の処理
        if (i >> j) & 1 == 0:
            continue
        # 人2が本当ならそれぞれ証言と照らし合わせて
        for x, y in evidences[j]:
            # 010の場合x=1つまり人1がy=1に対してフラグは立ってないこれは間違い
            if (i >> x) & 1 != y:
                consistent = False
                break
        if not consistent:
            break
    if consistent:
        # 2進数に変換して最初の0bを消す。フラグの数を数える
        result = max(result, bin(i)[2:].count('1'))
print(result)
# 000　全員嘘つき
# 001 アだけ
# 010 イだけ
# 011 アとイ
# 100 ウだけ
# 101 アとウ
# 110 イとウ
# 111 アイウ全員