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;
    int k[10], s[10][10], p[10];
    int onoff[10] = {};
    int ans = 0, onnum, oncheck;
    int i, j, l;
    scanf("%d %d", &N, &M);
    for(i = 0; i < M; i++){
        scanf("%d", &k[i]);
        for(j = 0; j < k[i]; j++){
            scanf("%d", &s[i][j]);
        }
    }
    for(i = 0; i < M; i++){
        scanf("%d", &p[i]);
    }
    for(i = 0; i < (int)pow(2, N); i++){
        oncheck = 1;
        for(j = 0; j < M; j++){
            onnum = 0;
            for(l = 0; l < k[j]; l++){
                onnum += onoff[s[j][l] - 1];
            }
            if(onnum % 2 != p[j]){
                oncheck = 0;
                break;
            }
        }
        if(oncheck == 1){
            ans++;
        }
        onoff[0]++;
        j = 0;
        while(onoff[j] == 2){
            onoff[j] = 0;
            onoff[++j]++;
        }
    }
    printf("%d\n", ans);
    return 0;
}