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.Scanner;

public class Main {

    public static void main(String[] args) {
	    // write your code here
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int m = sc.nextInt();

        // abc 128
        // bit全探索で考えます
        long[] swList = new long[m];
        int[] pList = new int[m];

        for (int i = 0; i < m; i++) {
            long sum = 0;
            int k = sc.nextInt();
            for (int j = 0; j < k; j++) {
                sum += (long)Math.pow(2, sc.nextInt() - 1);
            }
            swList[i] = sum;
        }

        for (int i = 0; i < m; i++) {
            pList[i] = sc.nextInt();
        }

        long ans = 0;

        parent : for (long i = 0; i <= Math.pow(2, n) - 1; i++) {
            for (int j = 0; j < m; j++) {
                if (Long.bitCount(i & swList[j]) % 2 != pList[j]) {
                    continue parent;
                }
            }
            ans++;
        }

        System.out.println(ans);
    }
}