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 bit_count(i, b)
  ret = 0
  until(i == 0) do
    ret += 1 if i[0] == b
    i >>= 1
  end
  ret
end

N = gets.to_i
H = []
U = []
N.times{|i|
  n = gets.to_i
  h = 0
  u = 0
  n.times{
    x, y = gets.chomp.split(" ").map(&:to_i)
    y == 1 ? (h |= 1 << (x - 1)) : (u |= 1 << (x - 1))
  }
  H << h; U << u
}

ans = 0
(1...2**N).each{|i|
  next if bit_count(i, 1) <= ans
  h, u = [0, 0]
  (1..N).each{|n|
    (h |= H[n - 1]; u |= U[n - 1]) if i[n - 1] == 1
  }
  ans = bit_count(i, 1) if (h & ~i == 0)  && (u & i == 0)
}

puts ans