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)

N = gets.to_i
全証言 = N.times.map{
	gets.to_i.times.inject([0,0]){|(正直,不正直),| # 自己矛盾はなし
		x, y = gets.split.map(&:to_i)
		next [正直|(y<<(x-1)), 不正直|((1-y)<<(x-1))]
	}
}
F = lambda{|_| b, c = _
	正直, 不正直 = (0...N).select{|i| b[i]==1 }.map{|i| 全証言[i] }.inject([0,0]){|(a,b),(c,d)| [a|c, b|d] }
	next 正直|b, 不正直|c if (正直|b)^b==0
	next F.call [正直|b, 不正直|c]
}
全証言 = 全証言.map(&F)
p (2**N).times.map{|b| F.call [b,0] }.select{|正直,不正直| 正直&不正直==0 }.map{|正直,| (0...N).count{|i| 正直[i]==1 } }.max