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 Ruby (2.3.3) (judge verdict: Wrong Answer)

# coding: utf-8
N = gets.to_i

# 人数分の評価記録リストを作成
evas = Array.new(N, nil)

# 評価の矛盾件数
c = 0

# 外側のループ：人数分
(1..N).each do |_|
  # 内側のループ：証言数
  num = gets.to_i
  num.times do |_|
    person, eva = gets.split.map {|e| e.to_i}
    # まだ評価されていない場合は評価かきこみ
    if evas[person-1] == nil
      evas[person-1] = eva
    else
      # 評価に矛盾がある場合は件数をカウント
      if evas[person-1] != eva
        c += 1
      end
    end
  end
end

# 評価に矛盾のない人数を出力
# ただし、すべての評価がもれなくX:0の場合は、少なくとも一人は正直である可能性がある
if evas.all? {|e| e == 0}
  puts 1
else
  puts N - c
end