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: Accepted)

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
#include<utility>
#include<functional>
#include<stack>
#include<queue>
#include<cmath>
# define MOD 1000000000+7
# define INF 1e9
# define LINF 1e18
typedef long long ll;

using namespace std;
int gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
 

bool broken[100000] = {false};
int dp[100010] = {};

int unko[200000] = {}, geri[200000] = {};

int main() {
	/*
		int n = 0, m = 0;
		cin >> n >> m;

		for (int i = 0; i < m; i++) {
			int temp;
			cin >> temp;
			broken[temp] = true;
		}

		dp[0] = 1;

		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j <= min(i + 2, n); j++) {

				if (broken[j] == false) {
					dp[j] += dp[i];
					dp[j] %= MOD;
				}

			}
		}

		cout << dp[n] << endl;

		return 0;
		*/

	int n, m, k;
	int ans = 0;
	cin >> n >> m;
	vector<vector<int>> s(m);
	vector<int> p(m);

	for (int i = 0; i < m; i++) {
		cin >> k;
		s[i].resize(k);//s[i]の大きさを変えてあげる

		for (int j = 0; j < k; j++) {
			cin >> s[i][j];
			s[i][j]--;
		}
	}

	for (int i = 0; i < m; i++) {
		cin >> p[i];
	}

	for (int bit = 0; bit < (1 << n); bit++) {//2^n列挙

		vector<int> on;

		for (int i = 0; i < n; i++) {
			if (bit & (1 << i)) {//bitの0bと1をiシフトした数の0bのANDをとる
				on.push_back(i);
			}
		}

		bool ok = true;

		for (int i = 0; i < m; i++) {
			int oncnt = 0;

			for (int j = 0; j < s[i].size(); j++) {
				for (int k = 0; k < on.size(); k++) {
					if (s[i][j] == on[k]) {

						oncnt++;
					}
				}
			}

			if (oncnt % 2 != p[i]) {
				ok = false;
				break;
			}
		}
		if (ok) {
			ans++;
		}
	}

	cout << ans << endl;

}