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>
#include <math.h>

int main(void) {
	int N, M;
	int base = 1;
	int i, j, l;

	scanf("%d %d", &N, &M);

	for (i = 0;i < N;i++) {
		base *= 2;
	}


	int k[M - 1];
	int a[M - 1][M - 1];
	int b[M - 1][M - 1];
	int s[M - 1];;
	int p[M];

	for (i = 0;i < M;i++) {

		s[i] = 0;
		scanf("%d", &k[i]);
		if (k[i] > N) {
			return 0;
		}
		for (j = 0;j < k[i];j++) {

			scanf("%d", &a[i][j]);
			if (a[i][j] > N) {
				return 0;
			}
			b[i][j] = 1;
			for (l = 0;l < a[i][j];l++) {
				b[i][j] *= 2;
			}

			s[i] += b[i][j];
		};

		s[i] /= 2;
		s[i] += base;
	}

	for (i = 0;i < M;i++) {

		scanf("%d", &p[i]);
	}

	int count = 0;
	int ssub[M - 1];

	for (j = 0;j < base;j++) {

		for (i = 0;i < M;) {
			int scount = 0;

			ssub[i] = s[i];

			ssub[i] -= j;
			for (l = 0;l < N + 1;l++) {
				scount += ssub[i] % 2;
				ssub[i] /= 2;
			}
			/*scount += N;
			scount++;*/

			scount %= 2;
			if (scount == p[i]) {
				i++;
			}
			else { break; }
		}

		if (i == M - 1) {
			count++;
		}
	}

	printf("%d\n", count);

	return 0;
}