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: Accepted)

#include <bits/stdc++.h>
using namespace std;
int main(void){
    int N,M;
    cin >> N >> M;
    int count = 0;
    vector<int> k(M);
    vector<int> p(M);
    vector<vector<int>> s(11,vector<int> (11,0));
    for (int i = 0; i < M; i++){
        cin >> k.at(i);
        for (int j = 0; j < k.at(i); j++){
            cin >> s.at(i).at(j);
        }
    }
    for (int i = 0; i < M; i++){
        cin >> p.at(i);
    }
    
    vector<int> temp(M);
    int check;
    vector<int> reallamp(N);
    int lamptemp;
    int maxlamp = 1;
    for (int i = 0; i < N; i++) maxlamp = maxlamp*2;
    for (int lamp = 0; lamp < maxlamp; lamp++){
        //初期化
        check = 0;
        for (int i = 0; i < M; i++){
            temp.at(i) = p.at(i);
        }
        lamptemp = lamp;
        
        //今のスイッチの状態
        for (int i = 0; i < N; i++){
            reallamp.at(i) = lamptemp % 2;
            lamptemp = lamptemp / 2;
        }
        
        for (int i = 0; i < M; i++){
            for (int j = 0; j < k.at(i); j++){
                temp.at(i) += reallamp.at( s.at(i).at(j) - 1 );
            }
        }
        
        for (int i = 0; i < M; i++){
            check += temp.at(i) % 2;
        }
        if (check == 0) {
            count++;
        }
    }
    
    printf("%d\n",count);

}