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<bits/stdc++.h>
using namespace std;
//Optimisations
#pragma GCC target ("avx2")
#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize("O2")
//shortcuts for functions
#define pb push_back
#define mp make_pair
#define ff first 
#define ss second
#define all(v) v.begin(),v.end()
#define prec(n) fixed<<setprecision(n)
#define n_l '\n'
// make it python 
#define gcd __gcd
#define append push_back
#define str to_string
// utility functions shortcuts
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define sswap(a,b) {a=a^b;b=a^b;a=a^b;}
#define swap(a,b) {auto temp=a; a=b; b=temp;}
#define init(dp) memset(dp,-1,sizeof(dp));
#define set0(dp) memset(dp,0,sizeof(dp));
#define bits(x) __builtin_popcount(x)
#define SORT(v) sort(all(v))
#define endl "\n"
#define forr(i,n) for(ll i=0;i<n;i++)
// declaration shortcuts
typedef long long int ll;
#define int ll
// Constants
constexpr int dx[] = {-1, 0, 1, 0, 1, 1, -1, -1};
constexpr int dy[] = {0, -1, 0, 1, 1, -1, 1, -1};
constexpr ll INF = 1999999999999999997; 
constexpr int inf= INT_MAX;
constexpr int MAXSIZE = int(1e6)+5;
constexpr auto PI  = 3.14159265358979323846L;
constexpr auto oo  = numeric_limits<int>::max() / 2 - 2;
constexpr auto eps = 1e-6;
constexpr auto mod = 1000000007;
constexpr auto MOD = 1000000007;
constexpr auto MOD9 = 1000000009;
constexpr auto maxn = 100006;
//void IOfile(){
//freopen(file_name, reade_mode, stdin);
//freopen(file_name, write_mode, stdout);
//}
void fastio(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
int32_t main(){
	fastio();
	int n,m;
	cin >> n >> m;
	vector<vector<int>>v;
	for(int i = 0; i < m; i++){
		int x;
		cin >> x;
		vector<int>c;
		for(int j = 0; j < x; j++){
			int k;
			cin >> k;
			c.push_back(k);
		}
		v.push_back(c);
	}
	int p[m];
	for(int i = 0; i < m; i++) cin >> p[i];
	int ans = 0;
	for(int i = 0; i < (1<< n); i++){
		int on[n];
		for(int j = 0; j < n; j++) on[j] = 0;
		for(int j = 0; j < n; j++){
			if(i & (1<<j)) on[j] = 1;
		}
		int c = 0;
		for(int j = 0; j < m; j++){
			int cnt = 0;
			for(int k = 0; k < v[j].size(); k++){
				cnt += on[v[j][k] - 1];
			}
			if(cnt % 2 == p[j]) c++;
		}
		if(c == m) ans++;
	}
	cout << ans;
}