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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] k = new int[m];
        int[][] s = new int[m][];
        for (int i = 0; i < m; i++) {
            k[i] = sc.nextInt();
            s[i] = new int[k[i]];
            for (int j = 0; j < k[i]; j++) {
                s[i][j] = sc.nextInt();
            }
        }
        int[] p = new int[m];
        for (int i = 0; i < m; i++) {
            p[i] = sc.nextInt();
        }
        sc.close();

        int ans = 0;
        for (int i = 0; i < pow(2, n); i++) {
            String target = toBinaryString(i);
            boolean isOk = true;
            for (int j = 0; j < m; j++) {
                int count = 0;
                for (int a = 0; a < k[j]; a++) {
                    if (target.charAt((target.length() - 1) - (s[j][a] - 1)) == '1') {
                        count++;
                    }
                }
                if (count % 2 != p[j]) {
                    isOk = false;
                    break;
                }
            }
            if (isOk) {
                ans++;
            }
        }
        System.out.println(ans);
    }

    // intを13桁のbit文字列に変換
    static String toBinaryString(int input) {
        String ret = "";
        int max = 0x00001000;
        for (int i = 0; i < 13; i++) {
            int tmp = input / max;
            ret += tmp;
            input %= max;
            max /= 2;
        }
        return ret;
    }

    // baseのn乗を計算を返す
    static int pow(int base, int n) {
        int ret = 1;
        for (int i = 0; i < n; i++) {
            ret *= base;
        }
        return ret;
    }

}