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.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int[] bit = new int [M];
		for(int i = 0; i < M; i++) {
			int k = sc.nextInt();
			for(int j = 0; j < k; j++) {
				int s = sc.nextInt();
				// 電球に対して繋がっているスイッチのフラグを立てる
				// 1を左にs-1ビット = スイッチの位置をbitの位置に置き換える
				// ORを取ることで接続されているスイッチの位置のみフラグを立てることができる
				bit[i] |= 1 << (s - 1);
			}
		}
		int[] p = new int[M];
		for(int i = 0; i < M; i++) {
			p[i] = sc.nextInt();
		}
		sc.close();

		//--test
		// System.out.println("----------");
		// for(int i = 0; i < M; i++) {
		// 	System.out.print(p[i] + ":" + Integer.toBinaryString(bit[i]));
		// 	System.out.println("");
		// }
		// System.out.println("----------");
		//--test

		int ans = 0;
		// スイッチの数の^2までのループ
		// ^2の理由：スイッチの数分bitに置き換えているから
		// 例)スイッチの数が5であれば2^5=16まで
		for(int i = 0; i < Math.pow(2, N); i++) {
			boolean ok = true;
			// 電球の数分のループ
			// スイッチiをONにしたときに電球すべての条件に該当するか判定する
			// (電球が点灯するp[j]と一致する場合、OK)
			for(int j = 0; j < M; j++) {
				// スイッチのONの数を取得
				int bitCnt = Integer.bitCount(bit[j] & i);
				// スイッチのONの数を2で割った余りが電球の点灯条件に一致しない場合
				// NGのパターンとなる
				if(bitCnt % 2 != p[j]) {
					// System.out.println("[NG]swith on > "+(i)+":"+
					// Integer.toBinaryString(bit[j])+"&"+(i)+"="+
					// Integer.toBinaryString(bit[j]&i)+" "+
					// "bitCnt="+Integer.bitCount(bit[j])+
					// " >> "+bitCnt+
					// " p[j]="+p[j]
					// );
					ok = false;
					break;
				}
				// System.out.println("[OK]swith on > "+(i)+":"+
				// Integer.toBinaryString(bit[j])+"&"+(i)+"="+
				// Integer.toBinaryString(bit[j]&i)+" "+
				// "bitCnt="+Integer.bitCount(bit[j])+
				// " >> "+bitCnt+
				// " p[j]="+p[j]
				// );
			}
			if(ok) { 
				ans++;
				// System.out.println(" >> anser count");
			}
		}
		System.out.println(ans);
	}
}