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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

typedef struct node {
    int light[10];
    struct node *next_0;
    struct node *next_1;
    int num;
} Node;

Node *make(int n, int now, int hoge, int *lis) {
    Node *a = (Node *)malloc(sizeof(Node));
    if (now == 0) {
        for (int i=0; i<n; i++) {
            a->light[i] = 0;
        }
    }
    else {
        for (int i=0; i<n; i++) {
            if (i == now-1) a->light[i] = hoge;
            else a->light[i] = lis[i];
        }
    }
    if (now < n) {
        a->next_0 = make(n, now+1, 0, a->light);
        a->next_1 = make(n, now+1, 1, a->light);
    }
    else {
        a->next_0 = a->next_1 = NULL;
    }
    a->num = now;
    return a;
}

int run(Node *start, int *ans, int *p, int **s_buf, int m, int n) {
    if (start == NULL) return;
    if (start->num == n) {
        int count = 0;
        for (int i=0; i<m; i++) {
            int buf = 0;
            for (int j=0; j<s_buf[i][0]; j++) {
                buf += start->light[s_buf[i][j+1]-1];
            }
            if (buf%2 == p[i]) count++;
            else break;
            if (count == m) {
                *(ans) += 1;
            }
        }
    }
    else {
        run(start->next_0, ans, p, s_buf, m, n);
        run(start->next_1, ans, p, s_buf, m, n);
    }
}

void free_node(Node *a) {
    if (a == NULL) return;
    free_node(a->next_0);
    free_node(a->next_1);
    free(a);
}

int main() {
    int n = 0, m = 0;
    scanf("%d %d\n", &n, &m);
    int p[m];
    int s[m][n+1];
    int *s_buf[m];
    for (int i=0; i<m; i++) {
        scanf("%d", &s[i][0]);
        for (int j=0; j<s[i][0]; j++) {
            scanf("%d", &s[i][1+j]);
        }
        s_buf[i] = s[i];
    }
    for (int i=0; i<m; i++) scanf("%d", &p[i]);
    int ans = 0;
    Node *start = make(n, 0, 0, p);
    run(start, &ans, p, s_buf, m, n);
    printf("%d\n", ans);
    free_node(start);
    return 0;
}