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;
public class ABC128C {
    public static void Main() {
        int N, M, ans = 0;
        List<int[]> S = new List<int[]>();
        List<int> P = new List<int>();
        string[] str = System.Console.ReadLine().Split(' ');
        N = int.Parse(str[0]);
        M = int.Parse(str[1]);
        int[] status = new int[N];
        for (int i = 0; i < M; i++) {
            str = System.Console.ReadLine().Split(' ');
            S.Add(new int[int.Parse(str[0])]);
            for (int j = 1; j < str.Length; j++) {
                S[i][j - 1] = int.Parse(str[j]);
            }
        }
        str = System.Console.ReadLine().Split(' ');
        for (int i = 0; i < M; i++) {
            P.Add(int.Parse(str[i]));
        }
        for (int p = 0; p < Math.Pow(2, N); p++) {
            bool ok = true;
            int temp = p;
            for (int i = 0; i < N; i++) {
                status[N - i - 1] = temp % 2;
                temp /= 2;
            }
            int sum = 0;
            for (int i = 0; i < M; i++) {
                for (int j = 0; j < S[i].Length; j++) {
                    sum += status[S[i][j] - 1];
                }

                if (sum % 2 != P[i]) {
                    ok = false;
                    break;
                }
            }
            if (ok) ans++;
        }
        System.Console.WriteLine(ans);
    }
}