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)

N = gets.chomp.to_i
A = []
N.times do
  n = gets.chomp.to_i
  a = {}
  n.times do
    x, y = gets.chomp.split(" ").map(&:to_i)
    a[x-1] = y
  end
  A << a
end

def check(a, liars)
  table = {}
  a.each do |list|
    list.each do |k, v|
      if table[k] != nil && table[k] != v then
        return nil
      end
      table[k] = v
    end
  end

  l = []
  table.each do |k, v|
    l << k if v == 0
  end
  p l
  p liars
  if l.sort != liars.sort then
    return nil
  end
  return l.size
end

result = check(A, [])
if result == 0 then
  puts N
  exit
end

all = (0...N).to_a

(1..N).each do |n|
  all.combination(n).each do |flips|
    a = A.clone
    flips.each do |idx|
      a[idx] = a[idx].map{ |k,v| [k, v == 1 ? 0 : 1]}.to_h
    end

    result = check(a, flips)
    if result != nil && result == n then
      puts N - n
      exit
    end
  end
end

puts "0"