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 <string>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <list>
#include <set>
#include <map>
#include <numeric>
#include <regex>
#include <tuple>
#include<iomanip>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
#define MOD 1000000007 // 10^9 + 7
#define INF 1000000000 // 10^9
#define LLINF 1LL<<60

bool p[10];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N, M; cin >> N >> M;
	vector<vector<int>> V(M);
	for (int i = 0; i < M; i++) {
		int num; cin >> num;
		for (int j = 0; j < num; j++) {
			int tmp; cin >> tmp;
			V[i].push_back(tmp);
		}
	}
	for (int i = 0; i < M; i++) cin >> p[i];

	int ans = 0;
	int up = 1;
	for (int i = 0; i < N; i++) up *= 2; // スイッチのパターンは全部で2^N
	for (int i = 0; i < up; i++) {
		vector<bool> swi; // light[k] : k番目のスイッチが入っている
		swi.push_back(0);
		int tmp = i;
		for (int i = 0; i < N; i++) {
			swi.push_back(tmp % 2);
			tmp /= 2;
		}
		//for(int i=0;i<)

		bool flag = true;
		for (int j = 0; j < M && flag; j++) {
			int cnt = 0;
			for (int k = 0; k < V[j].size(); k++) cnt += swi[V[j][k]];
			bool res = cnt % 2;
			if (p[j] != res) flag = false;
		}

		if (flag) ans++;
	}
	
	cout << ans << endl;

	return 0;
}