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)

#!/usr/bin/ruby

N = gets.chomp.to_i

$table = [
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
]

1.upto(N){|i|
  a_i = gets.chomp.to_i
  1.upto(a_i){|j|
    x_i_j,  y_i_j = gets.chomp.split(" ").map(&:to_i)
    $table[i][j] = [x_i_j, y_i_j]
    puts "table[#{i}][#{j}] = (#{x_i_j}, #{y_i_j})"
  }
}

# people = [dummy, 1,0,0...]
def isGood( people )
  i = 1;
  while i < people.length
    if people[i] == 1
      #puts"people[#{i}] = 1"
      1.upto($table[i].length-1){|j|
        #puts "#{i}, #{j}"
        t = $table[i][j]
        #puts "table[#{i}]= #{t.class}"
        #puts t[0]
        x, y = t
        #puts "x, people[x],y = #{x}, #{people[x]}, #{y}"
        return false if people[x] != y
      }

    end
    i+=1
  end

  return true
end

(2**N - 1).downto(0){|i|
  people = i.to_s(2).split("").map(&:to_i)
  while people.length != N
    people.unshift(0)
  end
  people.unshift("d")
  if true == isGood(people)
      puts people.count{|e| e == 1}
      exit
  end
}

puts 0