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 = Array.new(N+1)
XY = Array.new(N+1)
1.upto(N) do |x|
  A[x] = gets.chomp.to_i
  XY[x] = {}
  A[x].times do
    y, flag = gets.chomp.split(" ").map(&:to_i)
    XY[x][y] = flag
  end
end

require "set"

def mujun?(root)
  XY[root].each do |y, flag|
    if flag == 1 && XY[y][root] == 0
      return true
    end
  end
  false
end

def dfs(root, sj_base = Set.new, nj_base = Set.new)
  sj = Set.new + sj_base
  nj = Set.new + nj_base
  if mujun?(root) || nj.include?(root)
    nj << root
    return [sj, nj]
  end

  visited = sj + nj
  sj << root
  stack = [root]
  while !stack.empty?
    n = stack.pop
    next if visited.include?(n)
    visited << n
    next unless sj.include?(n)
    XY[n].each do |y, flag|
      if flag == 1 && !mujun?(y)
        sj << y
      else
        nj << y
      end
      stack << y
    end
  end
  [sj, nj]
end

ans = 0
1.upto(N) do |x|
  sj, nj = dfs(x)
  a = (sj & nj).size > 0 ? 0 : sj.size

  rest = (1..N).to_a - (sj | nj).to_a
  loop do
    break if rest.empty?
    i = rest.pop
    ssj, nnj = dfs(i, sj, nj)
    a += (ssj & nnj).size > 0 ? 0 : (ssj - sj).size
    sj += ssj
    nj += nnj
    rest -= (ssj | nnj).to_a
  end
  ans = a if a > ans
end

puts ans