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<string.h>
char univa[2000][11] = {};
char hogea[11] = {};
int univcnt = 0;
void makearray(int floor,char c,int invfloor){
  hogea[invfloor] = c;
  if (floor == 1){
    hogea[invfloor+1] = '\0';
    strcpy(univa[univcnt],hogea);
    //printf("%s\n",univa[univcnt]);
    univcnt++;
  }else{
    makearray(floor-1,'0',invfloor+1);
    makearray(floor-1,'1',invfloor+1);
  }
}

int main(){
  int n = 0;
  int m = 0;
  int k[10] = {};
  int ks[10][10] = {};
  int ksk = 0;
  int p[10] = {};
  int i = 0;
  int j = 0;
  int swstate[10] = {};
  int state = 0;
  int hogecnt;
  int cnt = 0;
  int succnt = 0;
  
  scanf("%d %d",&n,&m);
  for(i=0;i<m;i++){
    scanf("%d",&k[i]);
    for(j=0;j<k[i];j++){
      scanf("%d",&ksk);
      ks[i][ksk-1] = 1;
    }
  }
  for(i=0;i<m;i++){
    scanf("%d",&p[i]);
  }
  makearray(n,'0',0);
  makearray(n,'1',0);
  for(state=0;state<univcnt;state++){ //スイッチ状態
    succnt = 0;
    //
    for(j=0;j<n;j++){
      swstate[j] = univa[state][j] - '0';
      //printf("%d",swstate[j]);
    }
    //
    //printf("//");
    for(i=0;i<m;i++){
      hogecnt = 0;
      for(j=0;j<n;j++){
        if(ks[i][j] == 1 && swstate[j] == 1){
          hogecnt++;
        }
      }
      
      if(hogecnt % 2 == p[i]){
        succnt++;
      }
    }
    if(succnt == m){
      cnt++;
    }
  }
  printf("%d",cnt);
          
          
    
    
  return 0;
}