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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define rep(i,n) for(int i = 0; i <(int)(n); i++)
using namespace std;

int main(){
  
  //N,Mが10以下なので全探索でも間に合う
  
  int N,M;
  cin >> N >> M;
  vector<vector<int>>S(12,vector<int>(12));
  vector<int>P(M);
  vector<int>K(M);
  int k;
  
  rep(i,M){
    cin >> K[i];
    int a = K[i];
    rep(j,a){
      cin >> S[i][j];
    }
  }
  
  rep(i,M)cin >> P[i];
  
  //bit全探索
  int ans = 0;
  
  for(int bit = 0; bit < (1 << N); bit++){
    //スイッチの組み合わせを全探索。bitが立っているところのスイッチが起動している
    bool flag = true;
    //M個の電球についてみていく
    rep(i,M){
      int num = 0;
      //i個目の電球とスイッチの条件(num % 2 == P[i])が一致している時だけ先に進む
      int k = K[i];
      rep(j,k){
        int now = S[i][j-1];
        if(bit & (1 << now))num++;
      }
      if(num % 2 != P[i]){
        flag = false;
        break;
      }
    }
    if(flag)ans++;
    
  }
  
  cout << ans << endl;
  
}