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>
#define rep(i,n) for(int i=0;i<(n);++i)
#define FOR(i,n,j) for(int i=(j);i<(n);++i)
#define rrep(i,n) for(int i=(n);i>=0;--i)
using namespace std;
using ll=long long;
using ld=long double;
#define ssort(n) sort((n).begin(),(n).end())
typedef pair<int,int> P;

int main()
{
    int N,M;
    vector<int> k;
    cin >> N >> M;
    vector<vector<int>> s;
    vector<int> p;
    k.resize(M);p.resize(M);
    s.resize(M,vector<int>(N));

    rep(i,M){
        cin >>k[i];
        rep(j,k[i]){
            cin>> s[i][j];
        }
    }
    rep(i,M)cin >> p[i];
    vector<vector<int>> s_onoff;
    s_onoff.resize(M,vector<int>(N));
    rep(i,M){
        rep(j,N){
            s_onoff[i][j]=-1;
        }
    }
    rep(i,M){
        rep(j,k[i]){
            s_onoff[i][s[i][j]-1]=0;
        }
    }
    queue<int> q_s;
    int num,val;
    int l;
    int count_patern=0;
    bool result=true;
    rep(i,pow(2,N)){
        num=i;
        while(num>=1){
            q_s.push(num%2);
            num/=2;
        }
        while(q_s.size()<N){
            q_s.push(0);
        }
        
        l=0;
        while(!q_s.empty()){
            val=q_s.front();
            q_s.pop();
            rep(j,M){
                if(s_onoff[j][l]!=-1){
                    s_onoff[j][l]=val;
                }
            }
            l++;
        }
        int count_on=0;
        rep(j,M){
            rep(o,N){
                if(s_onoff[j][o]==1){
                    count_on++;
                }
            }
            
            if(count_on%2!=p[j]){
                result=false;
                break;
            }
        }
        if(result){
            count_patern++;
        }else{
            result=true;
        }
    }
    cout << count_patern << endl;
}