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 (3.8.2) (judge verdict: Wrong Answer)

n,m = map(int,input().split())
klist = []
swlist = []
ans = 0
for s in range(m):
	k,*sw = list(map(int,input().split()))#kとswを別々に入力
	klist.append(k) #<- kのリスト
	swlist.append(sw) #<- swのリスト
pwlist = list(map(int,input().split())) #pwのリスト

for i in range(2**n):#0~2**nまでの整数をカウントする(整数→ビットでの全探索)
	check = True
	for lump in range(m):#m個の電球についてカウントする
		cnt = 0
		for j in range(klist[lump]):#各スイッチ内の右シフト数をカウント	
			if (i>>swlist[lump][j])&1==1:#右にjシフトする
				cnt += 1#整数iのビットに対し、klistのスイッチがいくつonかをカウントする
		if cnt%2 != pwlist[lump]:#一つのスイッチ列でのカウント終了につき、判定する
			check = False#pwと互換性が1つでもなければFalseとする
	if check == True:#最後までFalseにならなければ組み合わせがあるということになる
		ans += 1
print(ans)