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)

import math
import fractions
import copy
#以下てんぷら
def j(q):
    if q==1: print("YES")
    elif q == 0:print("NO")
    exit(0)
rem = pow(10,9)+7
"""
def ct(x,y):
    if (x>y):print("+")
    elif (x<y): print("-")
    else: print("?")
"""

def ip():
    return int(input())
def printrow(a):
    for i in range(len(a)):
        print(a[i])
def combinations(n,r):
    if n<r:return 0
    return math.factorial(n) // (math.factorial(n - r) * math.factorial(r))
def permutations(n,r):
    if n<r:return 0
    return math.factorial(n) // math.factorial(n - r)
def lcm(x, y):
    return (x * y) // fractions.gcd(x, y)


#n = ip()                                     #入力整数1つ
n,m= (int(i) for i in input().split())       #入力整数横2つ以上
#a = [int(i) for i in input().split()]        #入力整数配列
#a = input()                                  #入力文字列
#a = input().split()                          #入力文字配列
#n = ip()                      #入力セット(整数改行あり)(1/2)
#a=[ip() for i in range(n)]    #入力セット(整数改行あり)(2/2)
#a=[input() for i in range(n)]    #入力セット(整数改行あり)(2/2)
#jの変数はしようできないので注意
#全足しにsum変数使用はsum関数使用できないので注意
a =[]
for i in range(m):
    adder = [int(i)-1 for i in input().split()]
    adder.pop(0)
    a.append(adder)
sw =  [int(i) for i in input().split()]
ctsw = [0 for i in range(n)]
ctsw[0] = -1
suc = 0
while sum(ctsw)<n:
    p = 0
    ctsw[0]+=1
    if ctsw[p]==2:
        while ctsw[p] == 2:
            ctsw[p] = 0
            p+=1
            ctsw[p]+=1
    for i in range(m):
        use = a[i]
        total = 0
        for k in range(len(use)):
            d = use[k]
            total+=ctsw[d]
        if total%2 != sw[i]: break
        if total%2 == sw[i] and i == m-1:suc+=1
print(suc)