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)

# スイッチONとOFFで、電球と繋がっている繋がっていないも、0or1なので、ビット演算を使用するのが吉

# スイッチと電球の繋がりをビット列で表す
# スイッチ(行）と電球(列)のマトリクスを作って、繋がっているところに1、繋がっていないところに0
# 行方向に見るとスイッチがどの電球に繋がっているかのビット列ができる
# intで持ってもよいので、100 は　4　

# 電球が点いているか点いていないかをビット列で表す
# スイッチを押すたびに、1つ前のスイッチが繋がっている電球にそのスイッチも繋がっていれば、偶奇が変化してスイッチの点滅が変化する
# スイッチを押すたびにxorをとる
# 電球の点滅とpが一致していればok


n, m = map(int, input().split())

# kはスイッチ、sはスイッチが繋がっている電球を表す
k = []
s = []

for _ in range(m):
    tmp = tuple(map(int, input().split()))
    k.append(tmp[0])
    s.append(set(tmp[1:tmp[0] + 1]))
 
p = tuple(map(int, input().split()))
 
# print(k, s, p, sep="\n")

ans = 0

# スイッチの押し方を全部試す
for i in range(2 ** n):
    # 各iでonになっている電球の数
    light_on = 0
 
    # 電球jについて
    for j in range(m):
        switch_on = 0
 
        for k in range(n):
            # onになっているスイッチ（kビット目が立っているか）
            if (i >> k) & 1:
                # 電球jとつながっているものならカウント
                # sはスイッチkが繋がっている電球jのリスト
                # スイッチは1からなので、k+1
                if k + 1 in s[j]:
                    switch_on += 1
 
        # onになっているスイッチの個数がp[j]に等しいなら電球jは点灯している
        if switch_on % 2 == p[j]:
            light_on += 1
 
    # 全ての電球が点灯しているならカウント
    if light_on == m:
        # print(bin(i))
        ans += 1
 
print(ans)