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>

int main(){
   int i,j,n,m,k[10],kk,l,s[10][10],ok[2048],nmax,b[11]={1,2,4,8,16,32,64,128,256,512,1024},p[10];
    
   scanf("%d %d",&n,&m);
   nmax=1;
   for (i=0;i<n;i++) nmax*=2;
   for (i=0;i<nmax;i++) ok[i]=1;
   for (i=0;i<m;i++) {
       scanf("%d",&k[i]);
       for (j=0;j<k[i];j++) {
          scanf("%d",&kk);
          s[j][i]=kk-1;
       }
   }
   for (i=0;i<m;i++) scanf("%d",&p[i]);
   for (i=0;i<m;i++)
       for (l=0;l<nmax;l++) 
          if (ok[l]) {
             kk=0;
             for (j=0;j<k[i];j++)
                 if ((l & b[s[j][i]])!=0) kk++;
             if (kk%2!=p[i]) ok[l]=0;
          }
   for (i=kk=0;i<nmax;i++)
      if (ok[i]) kk++;
   printf("%d\n",kk);
   return 0;
}