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 Python (3.4.3) (judge verdict: Accepted)

#!/usr/bin/env python3

import sys
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import product                # product(iter, repeat=n)
# from itertools import accumulate             # accumulate(iter[, f])
# from functools import reduce                 # reduce(f, iter[, init])
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from fractions import gcd                    # for Python3.4.3


def main():
    mod = 10000007                  # 10^9+7
    inf = float('inf')
    input = sys.stdin.readline      # 改行文字が残ることに注意
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def ii():  return int(input())
    def mi():  return map(int, input().rstrip().split())
    def lmi(): return list(map(int, input().rstrip().split()))
    def li():  return list(input().rstrip())
    

    def illuminate_p(state, bulb, L, P):
        connected_count = 0
        mask = 0b1
        for switch in L[bulb]:
            if state & (mask << switch):
                connected_count += 1
        return connected_count % 2 == P[bulb]
    

    n, m = mi()
    L = [None] * m
    # L[i] に電球 i のつながっているスイッチのリストが入る
    for i in range(m):
        line = list(map(lambda x: int(x)-1, input().rstrip().split()))
        L[i] = line[1:]
    P = lmi()

    ans = 0
    # state の下位 [j] bit が [j] 番目の switch が on かどうかのフラグとなる
    for state in range(2**n):
        for bulb in range(m):
            if not illuminate_p(state, bulb, L, P):
                break
        else:
            ans += 1
    print(ans)



if __name__ == "__main__":
    main()