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<string.h>
#include<math.h>

#define MAX(p,q) ((p>q)?(p):(q))
#define MIN(p,q) ((p<q)?(p):(q))



#define REP(i,n) for(i=0;i<n;i++)

int gcd(int a,int b);

int main(void)
{
  int n,m;
  scanf("%d%d",&n,&m);
  int k[10], s[10][10], p[10];
  int i,j;
  REP(i,m)
  {
    scanf("%d",&k[i]);
    REP(j,k[i])
    {
      scanf("%d",&s[i][j]);
      //printf("%d\n",s[i][j]);
    }
  }
  REP(i,m) scanf("%d",&p[i]);
  int flag,q,ans = 0,miniflag;

  int light[10] = {0};
  REP(i,1<<n)
  {
    //printf("%d\n",i);
    flag = 0;
    REP(j,m)//j個の電球について調べる
    {
      miniflag = 0;
      REP(q,k[j])//１個の電球に関するki個のスイッチについて調べる。
      {
        //s[j][p]
        if((i>>(n-s[j][q]))&1)
        {
          //printf("\t%dはon!\n",s[j][q]);
          miniflag++;
        }
        //else printf("\t%dはoff:(\n",s[j][q]);
      }
      if(miniflag%2==p[j]) flag++;


      else break;
    }
    if(flag==m)
    {
      //printf("%dのとき全部ついた%dこめ\n",i,ans);
      ans++;

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

int gcd(int a,int b)
{
  int max,min,tmp;
  max = MAX(a,b);
  min = MIN(a,b);
  if(max%min==0) return min;
  else
  {
    tmp = min;
    min = max%min;
    max = tmp;
    return gcd(max,min);
  }
}