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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;

class C
{
	static void Main()
	{
		var n = int.Parse(Console.ReadLine());
		var ps = new int[n].Select(_ => int.Parse(Console.ReadLine()))
			.Select(a => new int[a].Select(i => Console.ReadLine().Split().Select(int.Parse).ToArray()).ToArray())
			.ToArray();

		var p2 = new int[n + 1];
		p2[0] = 1;
		for (int i = 1; i <= n; i++) p2[i] = 2 * p2[i - 1];

		var M = 0;
		for (int f = 0; f < p2[n]; f++) M = Math.Max(M, Assert(f, ps, p2));
		Console.WriteLine(M);
	}

	static int Assert(int f, int[][][] ps, int[] p2)
	{
		var c = 0;
		for (int i = 0; i < ps.Length; i++)
		{
			if ((f & p2[i]) == 0) continue;
			c++;
			foreach (var xy in ps[i])
				if (((f & p2[xy[0] - 1]) > 0 ? 1 : 0) != xy[1]) return 0;
		}
		return c;
	}
}