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;
    scanf("%d%d", &n, &m);
    int k[m][n+1];
    int p[m];
    for(int i = 0; i < m; i++){for(int j = 0; j < n+1; j++){k[i][j] = 0;}}
    for(int i = 0; i < m; i++){
        scanf("%d", &k[i][0]);
        for(int j = 1; j < n+1; j++){
            scanf("%d", &k[i][j]);
        }
    }
    for(int i = 0; i < m; i++){scanf("%d", &p[i]);}
    
    int flag;
    int ar[n];
    int cnt[m];
    int ans = 0;
    for (int i=0; i<pow(2,n); i++) {
	    for (int j=0; j<n; j++) {
			if ((1&i>>j) == 1){
			    ar[j] = 1;
			}else{
			    ar[j] = 0;
			}
		}
		
// 		
// 		for(int z = 0; z < n; z++){printf("%d ", ar[z]);}printf("\n");
// 		
		for(int q = 0; q < m; q++){cnt[q] = 0;}
		for(int r = 0; r < n; r++){
		  //  printf("ar：%d\n", ar[r]);
		    if(ar[r] == 0){continue;}
		    
		    else{
		        for(int l = 0; l < m; l++){
    		        for(int o = 1; o < n+1; o++){
    		            if(r+1 == k[l][o]){
    		                cnt[l]++;
    		                break;
    		            }
    		        }
		       }
		    }
		}
		
		flag = 1;
		for(int r = 0; r < m; r++){
		    if(cnt[r] % 2 != p[r]){
		        flag = 0;
		        break;
		    }
		}
		if(flag == 1){ans++;}
    }
    printf("%d\n", ans);
    
    return 0;
}