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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
namespace I_am_the_brown_coder{
    class ABC10000000{
        static void Main(){
            string[] n_m = Console.ReadLine().Split();
            int n = int.Parse(n_m[0]);
            int m = int.Parse(n_m[1]);
            List<int>[] sw = new List<int>[n];
            for(int i=0;i<n;i++){
                sw[i] = new List<int>();
            }
            for(int i=0;i<m;i++){
                string[] input = Console.ReadLine().Split();
                for(int j=0;j<int.Parse(input[0]);j++){
                    sw[int.Parse(input[j+1])-1].Add(i);
                }
            }
            string[] ps = Console.ReadLine().Split();
            int[] p = new int[m];
            for(int i=0;i<m;i++){
                p[i] = int.Parse(ps[i]);
            }
            bool[] dnonoff = new bool[m];
            for(int i =0;i<m;i++){
                if(p[i]==0){
                    dnonoff[i] = true;
                }
            }
            int ans = 0;
            Stack<int[]> st = new Stack<int[]>();
            st.Push(new int[]{0,0});
            st.Push(new int[]{0,1});
            while(st.Count>0){
                int[] ima = st.Pop();
                if(ima[1]>0){
                    foreach(int i in sw[ima[0]]){
                        dnonoff[i] = !dnonoff[i];
                    }
                }
                if(ima[0]==n-1){
                    bool allon = true;
                    for(int i=0;i<m;i++){
                        if(!dnonoff[i]){
                            allon = false;
                        }
                    }
                    if(allon){
                        ans += 1;
                    }
                    if(ima[1]>0){foreach(int i in sw[ima[0]]){
                        dnonoff[i] = !dnonoff[i];
                    }}
                }else{
                    st.Push(new int[]{ima[0]+1,0});
                    st.Push(new int[]{ima[0]+1,1});
                }
            }

            Console.WriteLine(ans);
        }

    }
}