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<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
using namespace std;
typedef long long ll;

const int INF = (1 << 28);
const int dx[4] = {0,1,0,-1}, dy[4] = {-1,0,1,0};
const int Dx[8] = {0,1,1,1,0,-1,-1,-1}, Dy[8] = {-1,-1,0,1,1,1,0,-1};
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define INF 2e9
#define ALL(v) v.begin(), v.end()
int n,m;
int k[10];
int s[10][10];
int p[10];
ll dfs(ll count,ll g,int jj){
    if(jj==n){
        ll ss[n];
        ll te=10;
        ll temp1=g;
        for(int i=0;i<n;i++){
            ss[i]=temp1%10;
            temp1=temp1/10;
        }
        for(int i=0;i<m;i++){
            ll a=0;
            for(int j=0;j<k[i];j++){
                if(ss[s[i][j]-1]==1){
                    a++;
                }
            }
            if(a%2!=p[i]) return 0;
        }
        return 1;
    }
    ll temp=count+g;
    ll f=dfs(count*10,temp,jj+1);
    ll b=dfs(count*10,g,jj+1);
    return (f+b);
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> n>>m;
    for(int i=0;i<n;i++) {
        cin>>k[i];
        for(int j=0;j<k[i];j++){
            cin>>s[i][j];
        }
    }
    for(int i=0;i<m;i++){
        cin>>p[i];
    }
    cout << dfs(1,0,0)<< "\n";
}