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)

#pragma region include
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <set>
#include <cstdio>
#define ALL(obj) (obj).begin(),(obj).end()
#define RALL(obj) (obj).rbegin(),(obj).rend()
#define REP(i, n) for(int i = 0; i < n; 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 MOD 1000000007
#define INF 1000000000
#define LLINF 4000000000000000000
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
#pragma endregion

//#define __DEBUG__
#ifdef  __DEBUG__
#define dump(x) cerr << #x << " = " << (x) << " [" << __LINE__ << ":" << __FUNCTION__ << "] " << endl;
// vector出力
template<typename T>
ostream& operator << (ostream& os, vector<T>& v) {
    os << "{";
    REP(i, (int)v.size()) { os << v[i] << (i < v.size() - 1 ? ", " : ""); }
    os << "}";
    return os;
}
// pair出力
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& p) {
    return os << "(" << p.first << ", " << p.second << ")";
}
// map出力
template<typename T, typename U>
ostream& operator << (ostream& os, map<T, U>& map_var) {
    os << "{";
    for (auto itr = map_var.begin(); itr != map_var.end(); itr++) {
        os << "(" << itr->first << ", " << itr->second << ")";
        itr++;
        if (itr != map_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
// set 出力
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
    os << "{";
    for (auto itr = set_var.begin(); itr != set_var.end(); itr++) {
        os << *itr;
        ++itr;
        if (itr != set_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
#endif

int main() {
    cin.tie(0); ios::sync_with_stdio(false);
    int N, M; cin >> N >> M;
    set<int> se;
    VI k(M), p(M);
    VVI s(M, VI(N));
    REP(i, M) {
        cin >> k[i];
        REP(j, k[i]) {
            cin >> s[i][j];
            s[i][j]--;
        }
    }
    REP(i, M) {
        cin >> p[i];
    }
    ll ans = 0;
    REP(on, 1 << N) {
        int cnt = 0;
        bool b = true;
        for (int j = 0; j < M; j++) {
            for (int i = 0; i < k[j]; i++) {
                if (on & (1 << s[j][i])) cnt++;
            }
            if (cnt % 2 != p[j]) {
                b = false;
                break;
            }
        }
        if (b) ans++;
    }
    cout << ans << endl;
    getchar(); getchar();
}