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: Wrong Answer)

"""
メモ用
ビット全探索
不親切な人の証言をすべてひっくり返し、
できたリスト(la)の中で人、親切or不親切がすべて一致する時の人数を求める。
というやり方でやったが、方針が間違っていた。
親切な人の証言がすべて一致するかどうかという方針でやればok。
"""

import copy
N = int(input())
l = [] # 証言リスト
lo = [] # 親切な人の数のリスト
m = [] # 証言の個数リスト
ans = [-1] * 2 ** N # 答え
cnt = 0 #親切な人の数
# ↓入力
for _ in range(N):
  n = []
  a = int(input())
  m.append(a)
  for _ in range(a):
    b = list(map(int, input().split()))
    n.append(b)
  l.append(n)
# ↑入力
for i in range(2 ** N):
  ansa = copy.deepcopy(ans)
  la = copy.deepcopy(l)
  for j in range(N):
    if ((i >> j) & 1):
      cnt += 1 #人jが親切だったときにcnt += 1
    for k in range(m[j]):
      if ((i >> j) & 1): # 人jが親切なら
        pass
      else: # 0と1を入れ替える
        if la[j][k][1] == 0:
          la[j][k][1] = 1
        else:
          la[j][k][1] = 0
# ↓リストlaを結合
  li = []
  for x in range(N):
    for j in range(m[x]):
      li.append(la[x][j])
# ↑リストlaを結合
  for x in range(sum(m)-1):
    for j in range(x+1, sum(m)):
      if li[x][0] == li[j][0]:
        if li[x][1] != li[j][1]:
          ansa[i] = 0 # 一致しなかった場合ゼロを返す
  if ansa == [-1] * 2 ** N: # ゼロを一つも返さなかった場合
    lo.append(cnt) # cntをリストに代入
  cnt = 0
if len(lo) == 0:
  print(0)
else:
  print(max(lo))