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;
typedef pair<ll, ll> P;
typedef pair<int, int> Pi;
typedef vector<ll> Vec;
typedef vector<int> Vi;
typedef vector<string> Vs;
typedef vector<vector<ll>> VV;

#define REP(i, a, b) for(ll i=(a); i<(b); i++)
#define rep(i, n) REP(i, 0, n)
const int INF=1e9;
const int MAX=100005;
const int MOD=1000000007;
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl;
#define ALL(v) v.begin(), v.end()
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define pb(x) push_back(x)
#define mp(a, b) make_pair(a,b)
#define Each(a,b) for(auto &a :b)
#define REPM(i, mp) for (auto i = mp.begin(); i != mp.end(); ++i)
#define dbgmap(mp) for (auto i = mp.begin(); i != mp.end(); ++i) { cout << i->first <<":"<<i->second << endl;}
#define sum(v) accumulate(ALL(v),0)
#define fi first
#define se second

template<typename T1, typename T2>
ostream &operator<<(ostream &s, const pair<T1, T2> &p) { return s<<"("<<p.first<<", "<<p.second<<")"; }

// vector
template<typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
    int len=v.size();
    for(int i=0; i<len; ++i) {
        s<<v[i];
        if(i<len-1) s<<"	";
    }
    return s;
}

// 2 dimentional vector
template<typename T>
ostream &operator<<(ostream &s, const vector<vector<T> > &vv) {
    int len=vv.size();
    for(int i=0; i<len; ++i) {
        s<<vv[i]<<endl;
    }
    return s;
}

ll ans=0;
ll n,m,k;
Vec swistatus ; //m番目の電球がついている合計値
VV swiconnect ; //i番目のスイッチにはmが所属
Vec p;

void dfs(ll i,ll flag, Vec &status){
    dbg(i);
    if(flag){
        Vec &target = swiconnect[i];
        rep(i,target.size()){
            status[target[i]]++;
        }
       // dbg(status);
    }
    if(i<n){
        dfs(i+1,0,status);
        dfs(i+1,1,status);
    }else{
        bool comp = true;
        //dbg(status);
        rep(i,status.size()){
            if(status[i]%2==p[i]){
                comp =false;
            }
        }
        if(comp){
            ans ++;
        }
    }
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin>>n>>m;
    swistatus.resize(m);
    swiconnect.resize(n+1);
    p.resize(m);
    rep(i,m){
        cin>>k;
        rep(j,k){
            ll tmp;
            cin>>tmp;
            swiconnect[tmp].pb(i);
        }
    }
    rep(i,m){
        cin>>p[i];
    }
    //dbg(swiconnect);
    //1番スイッチに電球0,2版スイッチに電球0,1.....

    dfs(1,0,swistatus);
    dfs(1,1,swistatus);
    cout<<ans<<endl;
    return  0;
}