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 <bits/stdc++.h>	
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define ld long double
#define pii pair <int, int>
#define sz(a) (int)(a.size()) 
#define resize(v) v.resize(unique(all(v)) - v.begin()); 
#define all(a) a.begin(), a.end()
#define forit(it, s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)
#define int long long

// random_device rd; mt19937 g(rd());

using namespace std;
 
void Fast_Read_Out()
{
    ios_base::sync_with_stdio(0);
    cin.tie(), cout.tie();
}
 
void Random()
{
    unsigned int seed;                                                                        
    asm("rdtsc" : "=A" (seed));
    srand(seed);        
}
 
unsigned int Time()
{
     unsigned int time = clock() / 1000.00;                           
     return time;
}
 
const int mod = (int)1e9 + 7;
 
int id[100][100];
int cur[100];
int ans;
int k[100], p[100];

main ()
{
	#ifdef JUDGE
		freopen("input.txt", "r", stdin);
	#endif		
	Fast_Read_Out();
	Random();   
	int n, m;
	cin >> n >> m;
	for(int i = 1; i <= m; i++) 
	{
		cin >> k[i];
		for(int j = 1; j <= k[i]; j++) cin >> id[i][j], id[i][j]--;
	}
	for(int i = 1; i <= m; i++) cin >> p[i];
	for(int mask = 0; mask <= (1 << n); mask++)
	{
		memset(cur, 0, sizeof cur);
		int cnt = 0;
		for(int j = 1; j <= m; j++)
		{
			for(int l = 1; l <= k[j]; l++) 
			{
				for(int i = 0; i < n; i++) if(id[j][l] == i)  cur[j] += ((mask >> i) & 1);
			}
		}
		for(int i = 1; i <= m; i++) if(p[i] == cur[i] % 2) cnt++;
		if(cnt == m) ans++;
	}
	cout << ans;
}