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 <iosfwd>
#include <iomanip>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <cassert>
#include <cctype>
#include <climits>
#include <vector>
#include <bitset>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <deque>
#include <string>
#include <list>
#include <iterator>
#include <sstream>
#include <complex>
#include <fstream>
#include <functional>
#include <numeric>
#include <utility>
#include <algorithm>
#include <assert.h>
#include <unordered_map>

using namespace std;
long long mod = 1e9 + 7;
const int N = 3e5 + 500;

long long n, m, u, v, k, t, q, a, h;

pair <string, int> rest[N];

pair < pair <string, int>, int> srt[N];
vector <int> switches[N];
bool parity[N];
int P[N];

int main() {
   ios::sync_with_stdio(false);cin.tie(0);
   cin >> n >> m;
   for (int i = 1; i <= m; ++i) {
      cin >> k;
      for (int j = 1; j <= k; ++j) {
         cin >> u;
         switches[u - 1].push_back(i);
      }
   }
   int sol = 0;
   for (int i = 1; i <= m; ++i) {
      cin >> P[i];
      P[i] %= 2;
   }
   for (int i = 0; i < (1 << n); ++i) {
      for (int j = 0; j < n; ++j) {
        if ((i & (1 << j))) {
           for (int k = 0; k < switches[j].size(); ++k) {
              parity[switches[j][k]] ^= 1;
           }
        }
      }
      
      bool valid = true;
      for (int j = 1; j <= m; ++j) {
         if (parity[j] != P[j]) {
            valid = false;
            //cout << i << " " << j << " " << parity[j] << " " << P[j] << endl;
            break;
         }
      }
      //cout << valid << endl;
      sol += valid;
      for (int i = 1; i <= m; ++i) parity[i] = false;
   }
   
   cout << sol << endl;
}