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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>


int main() {
	int i, j, k, n, t, m, h, w, g;
	int f = 0;
	int o[12];
	int s[12][12];
	int p[12];
	int ii[15];
	int nn[15] = { 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 };

	scanf("%d %d", &n, &m);

	for (i = 0;i < m;i++) {
		scanf("%d", &o[i]);
		for (j = 0;j < o[i];j++) {
			scanf("%d", &s[i][j]);
		}
	}

	for (i = 0;i < m;i++) {
		scanf("%d", &p[i]);
	}


	for (i = 0;;i++) {
		g = 0;
		for (j = 0;j < m;j++) {
			w = 0;
			for (k = 0;k < o[j];j++) {
				w = w + nn[s[j][k]-1];
			}
			if (p[j] == w % 2) {
				g++;
			}
		}
		if (g == m) {
			f++;
		}
		nn[0]++;
		for (j = 0;j < n-1;j++) {
			if (nn[j] >= 2) {
				nn[j] = 0;
				nn[j + 1]++;
			}
			if (nn[n] == 1) {
				h = 1;
				break;
			}
		}
		if (h == 1) {
			break;
		}
	}


	printf("%d", f);


	return 0;
}