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<math.h>

int main(){
    int n, m;
    scanf("%d%d", &n, &m);
    int k[m];
    int s[10][10];
    int p[m];
    for(int i=0; i<m; i++){
        scanf("%d", &k[i]);
        for(int j=0; j<k[i]; j++){
            scanf("%d", &s[i][j]);
        }
    }
    for(int i=0; i<m; i++){
        scanf("%d", &p[i]);
    }
    int bi = pow(2, n) - 1;
    int ans = 0;
    for(int i=bi; i>=0; i--){
        int buf = i;
        int f=1;
        /*printf("bit : ");
        for(int j=0; j<n; j++){
            printf("%d", buf%2);
            buf = buf/2;
        }
        printf("\n");*/
        for(int j=0; j<m; j++){
            int sum = 0;
            for(int l=0; l<k[j]; l++){
                int t = i;
                for(int q=1; q<s[j][l]; q++){
                    t = t/2;
                }
                if(t % 2){
                    //printf("%d ", (int)pow(2, s[j][l]-1));
                    sum++;
                }
            }
            //printf("sum : %d\n", sum);
            if(sum % 2 != p[j]){
                f = 0;
            }
        }
        //printf("\n");
        if(f){
            ans++;
        }
    }
    printf("%d", ans);

    return 0;
}