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)

#define _USE_MATH_DEFINES
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<string>
#include<set>
#include<map>
#include<stdbool.h>
#include<math.h>
#include<sstream>
#include<math.h>
#include<vector>
#include<functional>
#include<queue>
#include<utility>
#include<bitset>
typedef long long ll;
#define N (1000000000+7)
#define INF 1e16
using namespace std;
typedef pair<ll, ll> P;

int bit[20];
int p[20];
int n, m;

int dfs(int level, int mask) {
	if (level == n) {
		int num = 0;
		
		for (int i = 0;i < m;i++) {
			for (int j = 0;j < n;j++) {
				if ((bit[i]>>j)&1) {
					if ((mask>>j)&1) {
						num++;
					}
				}
			}
			
			if (num % 2 != p[i])return 0;
		}
		return 1;
	}
	else {
		return dfs(level + 1, mask + (1 << level)) + dfs(level + 1, mask);
	}
}


int main(void) {
	int k, s;
	//cout << (15 >> 0) << endl;
	cin >> n >> m;

	for (int i = 0;i < m;i++) {
		cin >> k;
		bit[i] = 1 << n;
		for (int j = 0;j < k;j++) {
			cin >> s;
			bit[i] += 1 << (s - 1);
		}
	}
	for (int i = 0;i < m;i++)cin >> p[i];
	cout << dfs(0, 1 << n) << endl;
	return 0;
}