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>

#define REP(i,n) for(int i=0;i<(int)(n);i++)
#define ALL(x) (x).begin(), (x).end()

using namespace std;

template <typename T> T &chmin(T &a, const T &b) {return a = min(a, b);}
template <typename T> T &chmax(T &a, const T &b) {return a = max(a, b);}

using ll = long long;
using ld = long double;

const int INF = 1e9;
const ld eps = 1e-9, pi = acos(-1.0);
bitset<10> connect[10];
int flag[10];

int main(){
  int n, m, tmp;
  cin >> n >> m;
  REP(i, m){
    int k; cin >> k;
    REP(j, k) { cin >> tmp; connect[i].set(j); }
  }
  int res = 0;
  REP(i, m) cin >> flag[i];
  REP(i, 1<<10) {
    bitset<10> target(i);
    bool isok=true;
    REP(j, m){
      if((target&connect[j]).count()%2 != flag[j]){ isok=false; break; }
    }
    if(isok) res++;
  }
  cout << res << endl;
  return 0;
}