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>


int main(){
    int i, j, k, l, m, n;
    int N, M, L;
    
    scanf("%d", &N);
    scanf("%d", &M);
    
    int *K = (int *)malloc(sizeof(int) * M);
    int *mat, **Mat;
    mat = (int *)malloc(sizeof(int) * (N+1)*M);
    Mat = (int **)malloc(sizeof(int*) * M);
    for(i = 0; i < M; i++){
        Mat[i] = &mat[i*(N+1)];
    }


    for(i = 0; i < M; i++){
        scanf("%d", &K[i]);
        for(j = 0; j < K[i]; j++){
            scanf("%d", &k);
            Mat[i][k-1] = 1;
        }
    }
    for(i = 0; i < M; i++){
        scanf("%d", &Mat[i][N]);
    }
    
    // printf("\n");
    // for(i = 0; i < M; i++){
    //     for(j = 0; j < N+1; j++){
    //         printf("%d ", Mat[i][j]);
    //     }
    //     printf("\n");
    // }
    
    L = 1 << N;
    int result = 0;
    int testN[11];
    for(i = 0; i < L; i++){

        testN[ 0] = i & 0b0000000001;
        testN[ 1] = i & 0b0000000010;
        testN[ 2] = i & 0b0000000100;
        testN[ 3] = i & 0b0000001000;
        testN[ 4] = i & 0b0000010000;
        testN[ 5] = i & 0b0000100000;
        testN[ 6] = i & 0b0001000000;
        testN[ 7] = i & 0b0010000000;
        testN[ 8] = i & 0b0100000000;
        testN[ 9] = i & 0b1000000000;
        testN[N] = 1;
        
        for(j = 0; j < M; j++){
            int test = 0;
            for(k = 0; k < N+1; k++){
                if(testN[k]==0) continue;
                if(Mat[j][k]==1) test++;
            }
            if((test&1)==1){
                j = -1;
                break;
            }
        }
        if(j==-1) continue;
        result++;
        // printf("ok(%d/%d): ",i , N);
        // for(k = 0; k < N+1; k++){
        //     printf("%d ", testN[k]);
        // }
        // printf("\n");
    }
    
    printf("%d\n", result);
    
    return 0;
}