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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define rep(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define len(v) ll(v.size())

template <class T>
void cout_vec(const vector<T> &vec){
  for(auto itr:vec) cout<<itr<<' ';
  cout<<'\n';
}

typedef pair<ll,ll> P;
const ll mod=1e9+7;

int n,m;
vector<int> k(20),p(20);
vector<vector<int> > s(20,vector<int>(20));
int ans=0;

void dfs(int now,vector<bool> used){
  if(now==n+1){
    bool flag=true;
    FOR(i,1,m+1){
      int cnt=0;
      FOR(j,1,k[i]+1){
        if(used[s[i][j]]) cnt++;
      }
      if(cnt%2!=p[i]%2){
        flag=false;
      }
    }
    if(flag) ans++;
    return ;
  }
  vector<bool> used1(used);
  vector<bool> used2(used);
  used1[now]=true;
  used2[now]=false;
  dfs(now+1,used1);
  dfs(now+1,used2);
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin>>n>>m;
    FOR(i,1,m+1){
      cin>>k[i];
      FOR(j,1,k[i]+1){
        cin>>s[i][j];
      }
    }
    FOR(i,1,m+1){
      cin>>p[i];
    }
    //cout_vec(p);
    /*
    FOR(i,1,m+1){
      cout<<k[i]<<'\n';
      FOR(j,1,k[i]+1){
        cout<<s[i][j]<<' ';
      }
      cout<<endl;
    }
    */
    vector<bool> used(20,false);
    dfs(1,used);
    cout<<ans<<endl;
}