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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;
import java.util.function.*;

final class Solver {
	static final Solver INSTANCE = new Solver();

	void solve(Supplier<String> sc) {
		int N = Integer.parseInt(sc.get());
		int M = Integer.parseInt(sc.get());
		BitSet[] S = new BitSet[M];
		for (int i = 0; i < M; i++) {
			int K = Integer.parseInt(sc.get());
			BitSet bs = S[i] = new BitSet();
			for (int j = 0; j < K; j++) {
				bs.set(Integer.parseInt(sc.get()) - 1);
			}
		}
		int[] P = new int[M];
		Arrays.setAll(P, $ -> Integer.parseInt(sc.get()));

		int count = 0;
		for (int i = 0; i < 1 << N; i++) {
			BitSet on = BitSet.valueOf(new long[]{i});
			int valve = 0;
			for (int j = 0; j < S.length; j++) {
				BitSet result = new BitSet();
				result.or(S[j]);
				result.and(on);
				if ((result.cardinality() % 2) == P[j]) valve += 1;
			}
			if (valve == M) count += 1;
		}
		System.out.println(count);
	}
}

class Main {
	public static void main(String... args) {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		Solver.INSTANCE.solve(new Supplier<String>() {
			StringTokenizer line;

			public String get() {
				while (line == null || !line.hasMoreTokens()) try {
					line = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new UncheckedIOException(e);
				}
				return line.nextToken();
			}
		});
	}
}