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;

class Main{
    public static void main(String[] args){
	Scanner sc = new Scanner(System.in);
	int N = sc.nextInt();
	int M = sc.nextInt();
	// Mの接続状況をビットで管理
	int[] connection = new int[M];
	for(int i = 0; i < M; i++){
	    int k = sc.nextInt();
	    int state = 0;
	    // i番目のビットが立っていたらiとつながっているということになる
	    for(int j = 0; j < k; j++){
		int s = sc.nextInt();
		state += (1 << (s - 1));
	    }
	    connection[i] = state;
	}
	int[] p = new int[M];
	for(int i = 0; i < M; i++){
	    p[i] = sc.nextInt();
	}

	//Check
	int ans = 0;
	for(int i = 0; i < (1 << N); i++){
	    boolean check = true;
	    for(int j = 0; j < M; j++){
		int count = Integer.bitCount(i&connection[j]);
		if(count % 2 != p[j]){
		    check = false;
		    break;
		}
	    }
	    if(check){
		ans++;
	    }
	}
	System.out.println(ans);
	
	
    }
}