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: Accepted)

def judge(list, a, x, y)
  arr = Array.new(a.length,-1)
  list.each{ |i|
    arr[i] = 1
  }
  bool = true
  list.each{|i|
    a[i].times{|t|
      if (arr[x[i][t]-1] == -1 && y[i][t] == 1) || (arr[x[i][t]-1] == 1 && y[i][t] == 0)
        bool = false
      end
    }
  }
  bool
end

n = gets.to_i
a = []
x = Array.new(n).map{Array.new}
y = Array.new(n).map{Array.new}
n.times{|i|
  a[i] = gets.to_i
  a[i].times{|t|
    x[i][t], y[i][t] = gets.chomp.split(" ").map(&:to_i)
  }
}

bit = 2**n-1
ans = 0
for i in 0..bit
  list = []
  keta = 0
  num = 0
  while i > 0
    if i%2 == 1
      list.push(keta)
      num += 1
    end
    i = i / 2
    keta += 1
  end
  if judge(list, a, x, y)
    ans = num > ans ? num : ans
  end
end

puts ans