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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include <time.h>
typedef long long ll;
typedef long double ld;
#define INF (1LL<<60)
#define MOD1 1000000007
/*swap 交換*/
void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a=c;}
/*2つのうち大きい数を返す*/
ll max2(ll a,ll b){return a>=b?a:b;}
/*2つのうち小さい数を返す*/
ll min2(ll a,ll b){return a>=b?b:a;}
/*絶対値*/
ll ABS(ll a){return a>=0?a:(-a);}
typedef struct{
    ll aa;
    ll bb;
}frequent;
// 小大
int compare(const void *a, const void *b){
    return *(ll *)a>*(ll *)b?1:*(ll *)a<*(ll *)b?-1:0;}
/*aのb進数時の桁数*/
ll digits(ll a, ll b){return a/b?1+digits(a/b,b):1;}
/*nをa進数にしたときの、i桁目の数*/
ll placenai(ll n, ll a, ll i){return i==1?n%a:placenai(n/a,a,i-1);}
int main(void){
    ll n,m,i,j,ans=0,l,q;
    scanf("%lld%lld",&n,&m);
    ll k[m],s[m][10],p[m],onoff[m];
    for(i=0;i<m;i++){
        scanf("%lld",&k[i]);
        for(j=0;j<k[i];j++){
            scanf("%lld",&s[i][j]);
        }
    }
    for(i=0;i<m;i++){
        scanf("%lld",&p[i]);
    }
    ll sw=pow(2,n);
    // printf("%lld %lld? ",sw,x);
    for(i=0;i<sw;i++){
        for(j=0;j<m;j++){
            onoff[j]=0;
        }
        for(j=1;j<=n;j++){
            if(placenai(i,2,j)==1){
                for(l=0;l<m;l++){
                    for(q=0;q<k[l];q++){
                        if(j==s[l][q]){
                            onoff[l]++;
                            break;
                        }
                    }
                }
                // // printf("//%lld %lld// ",i,j);
                // for(l=0;l<k[j-1];l++){
                //     printf("%lld ",s[j-1][l]-1);
                //     // onoff[s[j-1][l]-1]++;
                // }
            }
        }
        for(j=0;j<m;j++){
            // printf("%lld ",onoff[j]);
            if(onoff[j]%2!=p[j]) break;
        }
        // printf("\n");
        if(j==m){
            ans++;
            // printf("%lld ",i);
        }
    }
    printf("%lld\n",ans);
    return 0;
}