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: Accepted)

#include <stdio.h>
#include <stdlib.h>

int N, M;   //N個のスイッチとM個の電球
int k[11];  //k個のスイッチにつながってる
int s[11][11];  //k個のスイッチ=s1~sk
int p[11];  //s個のスイッチのうち、オンになっている数を2で割った余りがpの時、点灯

/* ビットの立っているフラグをついてる電球する。s個ずらす方法で比べて、つながってる電球のうちついているものがいくつかを調べる。その調べた個数を2で割ってpと比べる */
int judge(int bit)
{
    int cnt;
    for(int i=1; i<=M; i++) {
        cnt=0;
        for(int j=1; j<=k[i]; j++) {    //つながってるk個のスイッチに対して
            if(bit & (1<<(s[i][j]-1))) cnt++;   //bitの(s[i][j]-1)番目が1の時、(s[i][j])はスイッチオン
            // printf("%dの%d回目:%d\n", i, j, cnt);
        }
        if((cnt%2) != p[i]) return 0;
    }
    return 1;
}

int main()
{
    int total=0;
    scanf("%d %d", &N, &M);
    for(int i=1; i<=M; i++) {
        scanf("%d", &k[i]);
        for(int j=1; j<=k[i]; j++) {
            scanf("%d", &s[i][j]);
        }
    }
    for(int i=1; i<=M; i++) scanf("%d", &p[i]);

    for(int bit=0; bit<(1<<N); bit++) {
        if(judge(bit)) total++;
    }

    printf("%d", total);
}