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)

/**
    AUTHOR:JUNSH
    TIME:19/5/25
**/
#include<bits/stdc++.h>
#define rep(i,a,b) for(ll i=(a);i<=(b);++i)
#define rev(i,a,b) for(ll i=(a);i>=(b);--i)
using namespace std;
typedef long long ll;
inline ll read(){
    ll f=1,x=0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return f*x;
}
const ll inf=0x3f3f3f3f;
ll n,m;
const ll maxn=15;
ll a[maxn];
ll st[maxn];
ll solve(ll pos,ll bit){
    if(pos>n){
        for(int i=1;i<=m;++i){
            ll tmp=bit&a[i];
            ll res=0;
            while(tmp>0){
                if(tmp&1)res++;
                tmp/=2;
            }
            if((res&1)!=st[i])
                return 0;
        }
        return 1;
    }
    return solve(pos+1,bit)+solve(pos+1,bit|(1<<pos));
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin>>n>>m;
    for(ll i=1;i<=m;++i){
        ll k;
        cin>>k;
        ll bit=0;
        ll tmp;
        for(ll j=1;j<=k;++j){
            cin>>tmp;
            bit|=(1<<tmp);
        }
        a[i]=bit;
    }
    for(ll i=1;i<=m;++i){
        cin>>st[i];
    }
    ll ans=solve(1,0);
    cout<<ans<<endl;
}