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 Scala (2.11.7) (judge verdict: Wrong Answer)

object Main extends App {
	val sc = new java.util.Scanner(System.in)
	val n  = sc.nextInt
	
	val es = Seq.fill(n)(Seq.fill(sc.nextInt)((sc.nextInt - 1, sc.nextInt == 1)).toMap)
	
	println((es.zipWithIndex.filter{case (e, i) => e.toList.forall{case (x, y) => x != i && !y}}.map(_ => 1) ++
		(2 to n).filter{i: Int =>
			es.combinations(i).exists(_.combinations(2).forall{case Seq(e1, e2) => 
				e1.foldLeft(true) {case (z, (x, y)) => z && e2.get(x).map(y == _).getOrElse(true)} &&
				e2.foldLeft(true) {case (z, (x, y)) => z && e1.get(x).map(y == _).getOrElse(true)}
		})
	} ++ Seq(0)).max)
}