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 <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPS(i, x) for (int i = 1; i <= (int)(x); i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define FORS(i, m, n) for (int i = m; i <= n; i++)
#define INF 2e9
#define MOD (1000000007)
#define EPS 1e-9
#define ALL(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
typedef long long int lli;
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N, M;
    cin >> N >> M;
    vector<vector<int>> s;
    vector<int> k(M);
    vector<int> p(M);
    s.push_back(vector<int>());
    REP(i, M)
    {
        cin >> k[i];
        s.push_back(vector<int>());
        REP(j, k[i])
        {
            int p;
            cin >> p;
            p--;
            s[i].push_back(p);
        }
    }
    REP(i, M)
    {
        cin >> p[i];
    }
    int count = 0;
    for (int bit = 0; bit < (1 << N); bit++)
    {
        vector<int> st(N);
        REP(i, N)
        {
            st[i] = (bit & (1 << i)) ? 1 : 0;
        }
        bool flg = true;
        REP(i, M)
        {
            int tmp = 0;
            REP(j, k[i])
            {
                // cout << s[i][j] << "\n";
                tmp += st[s[i][j]];
            }
            if (tmp % 2 != p[i])
            {
                flg = false;
                break;
            }
        }
        if (flg)
            count++;
    }
    cout << count << "\n";
}