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)

// https://atcoder.jp/contests/abc128/tasks/abc128_c

#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <string>
#include <tuple>
#include <vector>
using namespace std;

typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef pair<string, int> PSI;
#define INF (1e9)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOREACH(x, a) for (auto &(x) : (a))
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()
// int gcd(long a, long b) { return b ? gcd(b, a % b) : a; }
// int lcm(long a, long b) { return a * b / gcd(a, b); }

int n, m;
vector<unsigned int> vi; // 入力値のビット列（例：1, 3なら101）
vector<unsigned int> vp; // ビット列の総和の偶数・奇数
bool check(unsigned int arg) {
    unsigned int ans = 0, cnt;
    bool flag = true;
    // スイッチが未接続なら点灯しない
    // if (arg == 0)
    //     return false;
    // 全部のスイッチで点灯するか確認
    REP(i, m) {
        cnt = __builtin_popcount(vi[i] & arg);
        if (cnt & 1 != vp[i]) {
            flag = false;
            break;
        }
    }
    return flag;
}

int dfs(int idx, unsigned int arg) {
    if (idx == n) {
        return check(arg);
    }
    int ans = dfs(idx + 1, arg);
    arg |= (1 << idx); // nビット目を1にする
    ans += dfs(idx + 1, arg);
    return ans;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    vi.resize(m);
    vp.resize(m);
    REP(i, m) {
        int k, l;
        cin >> k;
        // 入力値をビット列にする（例：1, 3なら101）
        unsigned int t = 0;
        REP(j, k) {
            cin >> l;
            t |= (1 << (--l));
        }
        vi[i] = t;
    }
    REP(i, m) { cin >> vp[i]; }
    cout << dfs(0, 0) << endl;
    return 0;
}