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 <map>
#include <algorithm>
#include <vector>
#include <math.h>

using namespace std;
using lli = long long int;

vector<lli> vec[10] = {};

lli p[10] = {};
lli swi[10] = {};

int main(void) {

	lli N, M;
	cin >> N >> M;

	for (int i = 0; i < M; i++) {
		lli K;
		cin >> K;
		for (int j = 0; j < K; j++) {
			lli s;
			cin >> s;
			vec[i].push_back(s);
		}
	}

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

	lli ans = 0;

	for (int i = 0; i < (int)pow(2, N); i++) {
		for (int j = 0; j < N; j++) {
			swi[j] = (i >> j) % 2;
		}

		lli light = 0;

		for (int j = 0; j < M; j++) {
			lli sum = 0;

			for (size_t k = 0; k < vec[j].size(); k++) {
				sum += swi[vec[j][k] - 1];
			}

			if (sum % 2 == p[j]) {
				light += 1;
			}

		}

		if (light == M) {
			ans++;
		}

	}

	cout << ans << endl;

	return 0;
}