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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
typedef long long ll;
#define rep(i,l,r) for(int i=l; i<(int)(r); i++)
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
int upcmpll(const void *a, const void *b){
    return (*(ll*)a == *(ll*)b ? 0 : (*(ll*)a > *(ll*)b ? 1 : -1));
}
void sort_u(ll *a, int n){ qsort(a, n, sizeof(ll), upcmpll); }

int k[12], s[12][12], p[12];

int main(){
    int n,m; scanf("%d%d", &n, &m);
    rep(i,0,m){
        scanf("%d", k+i);
        rep(j,0,k[i]){
            scanf("%d", &s[i][j]);
        }
    }
    rep(i,0,m) scanf("%d", p+i);

    int ans = 0;
    //ビット全探索　初期値は0 (ans=0あり)
    rep(bit, 0, (1 << n)) {
        int t_lmp = 0;
        rep(i, 0, m) {
            int tmp_sum = 0;
            rep(j, 0, k[i]){
                if(bit&(1<<s[i][j])) tmp_sum++;
            }
            if(tmp_sum%2 == p[i]) t_lmp++;
            else break;
        }
        if(t_lmp == m) ans++;
    }

    printf("%d\n", ans);
    return 0;
}