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.

Examples

Input

2 2
2 1 2
1 2
0 1


Output

1


Input

2 3
2 1 2
1 1
1 2
0 0 1


Output

0


Input

5 2
3 1 2 5
2 2 3
1 0


Output

8
import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
        int [][] allBit=allBit(n);
        /*for(int i=0;i<allBit.length;i++){
        for(int j=0;j<n;j++)System.out.print(allBit[i][j]);
            System.out.println();
        }*/
        ArrayList<int[]>po=new ArrayList<>();
        for(int i=0;i<m;i++){
            int k=sc.nextInt();
            int[]a=new int[k];
            for(int j=0;j<k;j++){
                a[j]=sc.nextInt();
            }
            po.add(a);
        }
        int[]b=new int[m];
        for(int i=0;i<m;i++){
            b[i]=sc.nextInt();
        }
        long count=0;
        for(int i=0;i<allBit.length;i++){
            boolean bo=true;
            for(int j=0;j<m;j++){
                int c=0;
                for(int l=0;l<po.get(j).length;l++){
                    c+=allBit[i][po.get(j)[l]-1];
                }
                if(c%2!=b[j]){bo=false;break;}
            }
            if(bo)count++;
        }
        System.out.println(count);
    }
    public static int[][] allBit(int n) {
		int[][]a=new int[1<<n][n];
        for(int i=0;i<(1<<n);i++){
            for(int j=0;j<n;j++){
                a[i][j]+=((1&i>>j)==1)?1:0;
            }

        }
		return a;
	}
}
