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: Accepted)

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

class Program
{

    public static int ReadInt()
    {
        return int.Parse(Console.ReadLine());
    }

    public static int[] ReadInts()
    {
        return Console.ReadLine().Split().Select(int.Parse).ToArray();
    }

    public static void Main()
    {
        var Input = ReadInts();
        var N = Input[0]; //スイッチの数
        var M = Input[1]; //電球の数

        var s = new List<int>[M]; //M個の要素を持つ配列のリスト

        //kとsを分離してsの配列をM個保存
        for(int i = 0; i<M; i++)
        {
            var input2 = ReadInts(); //k s1 s2 s3・・・sk
            var k = input2[0];
            s[i] = new List<int>();　//M個の要素を持つ配列
            for(int j = 0; j < k; j++)
            {
                s[i].Add(input2[j+1] - 1); //s1 - 1,s2 -1,・・・を詰めていく
            }
        }

        var p = ReadInts();
        var count = 0;
        for (int i = 0; i < Math.Pow(2, N); i++)　//すべての電球のonoffの組み合わせについて
        {
            var b = new bool[N]; //bool値の配列 iが2進数に変換された値が入る
            for(int j = 0; j < N; j++)
            {
                //boolの配列に値を格納する
                //1に対して0~N-1左シフトした2進数 = 0001 0010 0100 1000 ・・・　に対してiと論理積（同じ桁が1かつ1なら1それ以外は0）を取り、1の桁をtrueにしていく
                //するとiが二進数00000・・・（N個）、0000・・・1、0000・・・10、・・・に変換される。（配列bにtrue = 1 false = 0 として格納される。）
                //すなわち、00000・・・（N個）、0000・・・1、0000・・・10、・・・となり、0がオフ1がオンとすると、すべてのスイッチの状態が取得できる。

                b[j] = (i & (1 << j)) != 0;
            }
            for(int j = 0; j < M; j++) //すべての電球について
            {
                var c = 0;
                foreach(var item in s[j]) //この座標s[j]のうちonになっているスイッチは？
                {
                    if(b[item]) //座標itemのbがtrueならスイッチがon
                    {
                        c++;
                    }
                }

                if(c % 2 != p[j]) //on になっているスイッチの個数を2で割った余りがpiに等しい時に点灯
                {
                    count--;　//点灯しない場合はカウントマイナス
                    break;
                }

            }
            count++; //すべてついたとしてカウントする　（つかない場合はすでにマイナスされている）
        }

        Console.WriteLine(count);

    }
}