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 PyPy3 (2.4.0) (judge verdict: Accepted)

N, M = map(int, input().split())

#各電球がつながっているスイッチの数
K = []

#各電球がつながっているスイッチの集合のリスト
S = []

for _ in range(M):
  ks = list(map(int, input().split()))
  K.append(ks[0])
  S.append(set(ks[1:]))
  
#各電球が偶数点灯(0)か奇数点灯(1)か
P = list(map(int, input().split()))
  
#条件を満たす組み合わせの数
cnt = 0

#0～2**N-1 最大1023(2**10)
for i in range(2**N):
  #全部onなら1のまま、1つでもoffになったら0にして次のループへ
  nowflag = 1
  #今回にiにおいてonになっているスイッチの集合をつくる
  nowon = set()
  for k in range(1,N+1):
    if i%(2**k)>=2**(k-1): nowon.add(k)
  
  #スイッチと電球につながっているスイッチの積集合をとり、
  #その集合のlen%2==Pならフラグそのまま、違ったらフラグを落としてループ抜け
  for j in range(M):
    if len(nowon&S[j])%2!=P[j]:
      nowflag = 0
      break
      
  if nowflag == 1: cnt += 1
#ループ終わり

print(cnt)