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;
#define rep(i,m, n) for (int i = m; i < n; i++)
#define inf 1000000000
typedef long long ll;
int n, m;

vector<int> k(100), p(100);
vector<vector<int>> s(100, vector<int>(100));
ll sum = 0;

void dfs(int now,vector<int> sw,int f){
        if(now!=1){
                if(f==1){
                        sw[now - 1] = 1;
                }
                else{
                        sw[now - 1] = 0;
                }
        }
        bool ok=true;
        if(now==n+1){
                for(int i=0;i<m;i++){
                        int count = 0;
                        for(int j=0;j<k[i];j++){
                                if(sw[s[i][j]]==1){
                                        count++;
                                }
                        }
                        if(p[i]!=(count%2)){
                                return;
                        }
                }
                
                sum++;
                return;
        }
  		dfs(now+1, sw,1);
        dfs(now+1, sw,0);
}

int main(){
        cin >> n >> m;
        vector<int> swt(n+1,0);

        rep(i,0,m){
                cin >> k[i];
                rep(j,0,k[i]){
                        cin >> s[j][i];
                }
        }
        rep(i,0,m){
                cin >> p[i];

        }
        dfs(1, swt,0);

        cout << sum << endl;

        return 0;
}