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(void)
{

int i,j,n,m,num,k,s,ans=0,judge=0,on=1;

scanf("%d%d",&n,&m);

int bul[m][n];
int p[m];
num = (int)pow(2,n);
int a[num][n];


for(i=0;i<m;i++){
  for(j=0;j<n;j++)
    bul[i][j]=0;
}



for(i=0;i<num;i++){
        for(j=0;j<n;j++){
            if(i%(int)pow(2,j+1)>=pow(2,j))
                  a[i][n-1-j]=1;
            else
                  a[i][n-1-j]=0;
        }
}

for(i=0;i<m;i++){
  scanf("%d",&k);
  for(j=0;j<k;j++){
    scanf("%d",&s);
    bul[i][s-1]=1;
  }
}

for(i=0;i<m;i++)
  scanf("%d",&p[i]);

for(i=0;i<num;i++){
  for(j=0;j<m;j++){
    for(k=0;k<n;k++){
      judge+=bul[j][k]*a[i][k];
    }
    if(judge%2==p[j])
      on*=1;
    else
      on*=0;
    judge=0;
  }
  if(on==1)
    ans++;
  on=1;
}

printf("%d",ans);
return 0;
}