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++ (Clang 10.0.0) (judge verdict: Accepted)

//参考：https://qiita.com/DaikiSuyama/items/9dba963ba74396b9b7fd
//インクルード(アルファベット順)
#include <algorithm>     //sort,二分探索,など
#include <bitset>        //固定長bit集合
#include <cmath>         //pow,logなど
#include <complex>       //複素数
#include <deque>         //両端アクセスのキュー
#include <functional>    //sortのgreater
#include <iomanip>       //setprecision(浮動小数点の出力の誤差)
#include <iostream>      //入出力
#include <iterator>      //集合演算(積集合,和集合,差集合など)
#include <map>           //map(辞書)
#include <numeric>       //iota(整数列の生成),gcdとlcm(c++17)
#include <queue>         //キュー
#include <set>           //集合
#include <stack>         //スタック
#include <string>        //文字列
#include <unordered_map> //イテレータあるけど順序保持しないmap
#include <unordered_set> //イテレータあるけど順序保持しないset
#include <utility>       //pair
#include <vector>        //可変長配列

using namespace std;
typedef long long ll;

//マクロ
//forループ関係
//引数は、(ループ内変数,動く範囲)か(ループ内変数,始めの数,終わりの数)、のどちらか
//Dがついてないものはループ変数は1ずつインクリメントされ、Dがついてるものはループ変数は1ずつデクリメントされる
#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define REPD(i, n) for (ll i = n - 1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = a; i <= (ll)(b); i++)
#define FORD(i, a, b) for (ll i = a; i >= (ll)(b); i--)
//xにはvectorなどのコンテナ
#define ALL(x) (x).begin(), (x).end() //sortなどの引数を省略したい
#define SIZE(x) ((ll)(x).size())      //sizeをsize_tからllに直しておく
//定数
#define INF 1000000000000 //10^12:極めて大きい値,∞
#define MOD 1000000007    //10^9+7:合同式の法
#define MAXR 100000       //10^5:配列の最大のrange(素数列挙などで使用)
//略記
#define PB push_back //vectorヘの挿入
#define MP make_pair //pairのコンストラクタ
#define FIR first    //pairの一つ目の要素
#define SEC second   //pairの二つ目の要素

signed main()
{
    //入力の高速化用のコード
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M;
    cin >> N >> M;

    vector<vector<int>> S;
    REP(i, M)
    {
        int k;
        cin >> k;
        S.PB(vector<int>(k));
        REP(j, k)
        cin >> S[i][j];
    }

    vector<int> P(M);
    REP(i, M)
    cin >> P[i];

    int ans = 0;
    REP(i, 1 << N)
    {
        int count = 0;
        REP(j, M)
        {
            int on = 0;
            REP(k, SIZE(S[j]))
            {
                //オンになっているスイッチの数を数え上げる
                if (i >> (S[j][k] - 1) & 1)
                    on++;
            }
            //オンになっているスイッチの数と条件を照らし合わせて電球がつくかどうか判定
            if (on % 2 == P[j])
                count++;
        }

        //すべての電球がついているか確認
        if (count == M)
            ans++;
    }

    cout << ans << endl;
    return 0;
}