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 <math.h>

#define mod 10e9+7
#define ll long long int

int main(void){
    int n, m;
    scanf("%d %d", &n, &m);
    int k[m];
    int s[m][10];
    for(int i = 0;i < m;i++){
        for(int j = 0;j < 10;j++){
            s[i][j] = 0;
        }
    }
    int p[m];
    for(int i = 0;i < m;i++){
        scanf("%d", &k[i]);
        for(int j = 0;j < k[i];j++){
            int a;
            scanf("%d", &a);
            s[i][a-1] = 1;
        }
    }
    
    for(int i = 0;i < m;i++){
        scanf("%d", &p[i]);
    }
    int ans = 0;
    for(int s1 = 0;s1 <= 1;s1++){
    for(int s2 = 0;s2 <= 1;s2++){
    for(int s3 = 0;s3 <= 1;s3++){
    for(int s4 = 0;s4 <= 1;s4++){
    for(int s5 = 0;s5 <= 1;s5++){
    for(int s6 = 0;s6 <= 1;s6++){
    for(int s7 = 0;s7 <= 1;s7++){
    for(int s8 = 0;s8 <= 1;s8++){
    for(int s9 = 0;s9 <= 1;s9++){
    for(int s10 = 0;s10 <= 1;s10++){
        int count = 0;
        for(int i = 0;i < m;i++){
            if((s1*s[i][0]+s2*s[i][1]+s3*s[i][2]+s4*s[i][3]+s5*s[i][4]+s6*s[i][5]+s7*s[i][6]+s8*s[i][7]+s9*s[i][8]+s10*s[i][9])%2 == p[i])count++;
        }
        if(count == m)ans++;
    }
    }
    }
    }
    }
    }
    }
    }
    }
    }
    int div = (int)pow(2, 10-n);
    printf("%d\n", ans/div);
    return 0;
}