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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#include <math.h>

int main(void)
{
	int n, m;
	int k[10];
	int s[10][10];
	int sw[10] = {0};
	int p[10];
	int sum = 1;
	int count = 0;
	int i, j, l, o;
	int work;
	int sum_sw;
	int flg;
	
	scanf("%d%d", &n, &m);
	
	for(i = 0; i < m; i++) {
		scanf("%d", &k[i]);
		for(j = 0; j < k[i]; j++) {
			scanf("%d", &s[i][j]);
		}
	}
	
	for(i = 0; i < m; i++) {
		scanf("%d", &p[i]);
	}
	
	for(i = 0; i < n; i++) {
		sum *= 2;
	}
	
	for(i = 0; i < sum; i++) {
		work = i;
		for(j = 0; j < 10; j++) {
			sw[j] = 0;
		}
		for(j = 0; work > 0; j++) {
			if( (work % 2) == 1) {
				sw[j] = 1;
			} else {
				sw[j] = 0;
			}
			work /= 2;
		}
		flg = 0;
		for(l = 0; l < m; l++) {
			sum_sw = 0;
			for(o = 0; o < k[l]; o++) {
				sum_sw += sw[s[l][o] - 1];
			}
			if(sum_sw % 2 != p[l]) {
				flg = 1;
				break;
			}
		}
		if(flg) {
			continue;
		}
		count++;
	}
	
	printf("%d\n", count);
	
	return 0;
}