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, N, M, k[11], s[11][11], p[11];
	scanf("%d %d", &N, &M);
	for (i = 1; i <= M; i++) {
		scanf("%d", &(k[i]));
		for (j = 1; j <= k[i]; j++) scanf("%d", &(s[i][j]));
	}
	for (i = 1; i <= M; i++) scanf("%d", &(p[i]));
	
	int count = 0, flag[11] = {}, tmp;
	while (1) {
		for (i = 1; i <= M; i++) {
			tmp = 0;
			for (j = 1; j <= k[i]; j++) if (flag[s[i][j]] == 1) {tmp ^= 1;
			if (tmp != p[i]) break;
                                                                }
		}
		
		if (i > M) count++;
		for (i = 1; i <= N; i++) if (flag[i] == 0) break;
		if (i > N) break;
		flag[i] = 1;
		for (i--; i >= 1; i--) flag[i] = 0;
	}
	
	printf("%d\n", count);
	fflush(stdout);
	return 0;
}