Problem:
Score : 300 points

Problem Statement

We have N switches with "on" and "off" state, and M bulbs. The switches are numbered 1 to N, and the bulbs are numbered 1 to M.

Bulb i is connected to k_i switches: Switch s_{i1}, s_{i2}, ..., and s_{ik_i}. It is lighted when the number of switches that are "on" among these switches is congruent to p_i modulo 2.

How many combinations of "on" and "off" states of the switches light all the bulbs?

Constraints

1 \leq N, M \leq 10

1 \leq k_i \leq N

1 \leq s_{ij} \leq N

s_{ia} \neq s_{ib} (a \neq b)

p_i is 0 or 1.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
k_1 s_{11} s_{12} ... s_{1k_1}
:
k_M s_{M1} s_{M2} ... s_{Mk_M}
p_1 p_2 ... p_M

Output

Print the number of combinations of "on" and "off" states of the switches that light all the bulbs.

Sample Input 1

2 2
2 1 2
1 2
0 1

Sample Output 1

1

Bulb 1 is lighted when there is an even number of switches that are "on" among the following: Switch 1 and 2.

Bulb 2 is lighted when there is an odd number of switches that are "on" among the following: Switch 2.

There are four possible combinations of states of (Switch 1, Switch 2): (on, on), (on, off), (off, on) and (off, off). Among them, only (on, on) lights all the bulbs, so we should print 1.

Sample Input 2

2 3
2 1 2
1 1
1 2
0 0 1

Sample Output 2

0

Bulb 1 is lighted when there is an even number of switches that are "on" among the following: Switch 1 and 2.

Bulb 2 is lighted when there is an even number of switches that are "on" among the following: Switch 1.

Bulb 3 is lighted when there is an odd number of switches that are "on" among the following: Switch 2.

Switch 1 has to be "off" to light Bulb 2 and Switch 2 has to be "on" to light Bulb 3, but then Bulb 1 will not be lighted. Thus, there are no combinations of states of the switches that light all the bulbs, so we should print 0.

Sample Input 3

5 2
3 1 2 5
2 2 3
1 0

Sample Output 3

8

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

public class MainClass
{
	private static int Cnt=0;
	private static Dictionary<String,int> Ora=new Dictionary<String,int>();
	public static void Main(string[] args)
	{
		string[] k = Console.ReadLine().Split(' ');
		int n = int.Parse(k[0]);
		int m = int.Parse(k[1]);
		List<Bolb> bolbs=new List<Bolb>();
		bool[] sw=new bool[n];
		
		for(int i = 0; i < m; i ++)
		{
			string[] s = Console.ReadLine().Split(' ');
			Bolb b = new Bolb();
			for(int j = 0; j < s.Length;j++)
			{
				int num=int.Parse(s[j]);
				if(j==0)
				{
					b.Connecting=num;
					continue;
				}
				b.Switches.Add(num-1);
			}
			bolbs.Add(b);
			
		}
		string[] t = Console.ReadLine().Split(' ');
		for(int i = 0; i < t.Length;i++)
		{
			bolbs[i].Amari=int.Parse(t[i]);
		}
		
		
		Sousa(0,bolbs,sw);
		Console.WriteLine(Cnt);
	}
	
	private static void Sousa(int i, List<Bolb> bolbs, bool[] sw)
	{
		if(i>sw.Length)
			return;
		string str=ToStr(sw);
		if(!Ora.ContainsKey(str))
		{
		Console.WriteLine(str);
		Ora.Add(str,1);
		bool p=true;
		foreach(Bolb k in bolbs)
		{
			if(!k.IsLight(sw))
			{
				p=false;
				break;
			}
		}
		
		if(p)
		{
			Cnt++;
		}
		}
				if(i>=sw.Length)
			return;

			sw[i]=false;
			Sousa(i+1,bolbs,sw);
			sw[i]=true;
			Sousa(i+1,bolbs,sw);
		
		
	}

	private static string ToStr(bool[] sw)
	{
		StringBuilder sb = new StringBuilder();
		for(int i = 0; i < sw.Length;i++)
		{
			sb.Append((int)(sw[i]?(1):(0)));
		}
		return sb.ToString();
	}
	
	public static int Clamp(int val, int max, int min = 0)
	{
		return Math.Min(max, Math.Max(min, val));
	}

	
}

public class Bolb
{
	public int Index{get;set;}
	public int Connecting{get;set;}
	public List<int> Switches{get;set;}
	public int Amari{get;set;}
	
	public Bolb()
	{
		Switches=new List<int>();
	}
	
	public bool IsLight(bool[] sw)
	{
		int knt=0;
		for(int i = 0; i < Switches.Count;i++)
		{
			if(sw[Switches[i]])knt++;
		}
		
		return (knt%2==Amari);
	}
}