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)

#include <bits/stdc++.h>
using namespace std;
 
int main() {
  int N, M;
  cin >> N >> M;

  int s[N] = {0};
  int d[M][N] = {0};
  int p[M];
  int k;
  int ins;
  int count = 0;
  int sum;
  int tmp;
  int flug;
  int max;

  for(int i=0;i<M;i++){
    cin >> k;
    for(int j=0;j<k;j++){
      cin >> ins;
      d[i][ins] = 1;
    }
  }

  for(int i=0;i<M;i++){
    cin >> p[i];
  }
  max = pow(2,N);

  for(int i=0;i<max;i++){
    tmp = i;
    flug = 1;
    for(int j=0;j<N;j++){
      s[j] = tmp % 2;
      tmp /= 2;

    }

    for(int j=0;j<M;j++){
      sum = 0;
      for(int k=0;k<N;k++){
        if(d[j][k] == 1){
          sum += s[k];
        }
      }
      if(sum%2 != p[j]) {
        flug = 0;
        break;
      }

    }

    if(flug == 1) count++;
  }

  cout << count << endl;

}