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.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		List<Set<Integer>> list = new ArrayList<Set<Integer>>();
		for (int i = 0; i < M; i++) {
			int k = sc.nextInt();
			Set<Integer> set = new HashSet<Integer>();
			for (int j = 0; j < k; j++) {
				set.add(sc.nextInt());
			}
			list.add(set);
		}
		int p[] = new int[M];
		for (int i = 0; i < M; i++) {
			p[i] = sc.nextInt();
		}
		int pattern = (int) (Math.pow(2, N));
		int result = 0;
		for (int i = 0; i < pattern; i++) {
			if (Main.check(i, list, p) == true) {
				result++;
			}
		}
		System.out.print(result);
	}

	public static boolean check(int pattern, List<Set<Integer>> list, int p[]) {
		boolean b = true;
		for (int i = 0; i < p.length; i++) {
			Set<Integer> set = list.get(i);
			int num = 1;
			int count = 0;
			int temp = pattern;
			while (temp > 0) {
				if (temp % 2 == 1) {
					if (set.contains(num) == true) {
						count++;
					}
				}
				temp /= 2;
				num++;
			}
			if (count % 2 != p[i]) {
				return false;
			}
		}
		return true;

	}
}