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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb pushback
#define fr(i,n) for(int i=0;i<n;i++)
#define ifr(i,n) for(int i=n-1;i>=0;i--)


int main() {
    int n,m;
    int x;
    int k;
	cin >> n>> m;
    int sw[m][n];
    int jou[m];
    int ans=0;
    bool flag;
    int nishin[n];
    fr(i,m){
      fr(j,n){
        sw[m][n]=0;
      }
    }
    fr(i,m){
      cin >> k;
      fr(j,k){
        cin >> x;
        sw[i][x]=1;
      }
    }
    int p[m];
    fr(i,m){
      cin>>p[i];
    }
    fr(i,int(pow(2,n))){
      int z = i;
      flag=true;
      fr(j,m){
       jou[j]=0;
      }
      
      fr(j,n){
        nishin[j]=z%2;
        z = z/2;
      }
      fr(j,n){
        fr(k,m){
          jou[k]+=nishin[j]*sw[k][j];
        }
      }
      fr(j,m){
        if(jou[j] % 2 != p[j])flag=false;
      }
      if(flag==true)ans+=1;
    }
      cout << ans << endl;
}