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<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int n,m,ans=0;//n-switch,m-lit
struct lit
{
	int num;
	int ask[10];
	int ode;//odevity
	int cnt=0;
};
int fun(int x[])
{
	int i;
	for(i=1;i<=n;i++)
		if(x[i]!=1)
		{
			x[i]=1;
			return 0;
		}
		else
			x[i]=0;
			return 0;
}
int main(){
	//输入 
	cin>>n>>m;
	int x[n+7];memset(x,0,sizeof(x));
	lit b[m+7];
	for(int i=1;i<=m;i++)
	{
		cin>>b[i].num;
		for(int j=1;j<=b[i].num;j++)
		{
			cin>>b[i].ask[j];
		}
	}
	for(int i=1;i<=m;i++)
		cin>>b[i].ode;
	//功能 
	for(int i=1;i<=pow(2,n);i++)//开关组合数 
	{
		fun(x);
		for(int j=1;j<=m;j++)//灯数 
		{
			b[j].cnt=0;
			for(int k=1;k<=b[j].num;k++)//每灯关联开关数 
			{
				if(x[b[j].ask[k]]==1)
				{
					b[j].cnt++;
					continue;
				}
			}
			if(b[j].cnt%2==b[j].ode)
			{
				if(j<m) continue;
				else ans++;break;
			}
			else break;
		}
	}
	cout<<ans;
	return 0;
}