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 Ruby (2.3.3) (judge verdict: Accepted)

=begin
－－－－－－－－－－－－－－
n = スイッチの数
m = 電球の数
k = 電球[i]に繋がっているスイッチ
a = スイッチを表す配列 (N個分)
pi = スイッチの最終状態
t = スイッチの押しパターンごとの電球の状態を記録
ans = 答えの数
－－－－－－－－－－－－－－
1: 各スイッチにどの電球が繋がっているのかを記録する
  1.1: i 番目の電球に繋がっている スイッチ k を受け取り、k をループする
    1.1.1: k 番目のスイッチ(a[k]) が繋がっている 電球[i]を 2進数で記録する (スイッチ k[j] には複数の 電球m[i] が繋がっている可能性のあるため ビットOR で受け取る)
2: スイッチの最終状態 pi を 2 進数で 受け取る
3: スイッチの押しパターンを全て試す
  3.1: パターンごとに 電球の状態 t を記録し、スイッチの最終状態 pi と一致すれば ans += 1
4: ans を出力
=end

n, m = gets.split.map(&:to_i)
a = Array.new(n,0)
pi = 0
ans = 0

for i in 0..(m-1) do
  k = gets.split.map{|v| v.to_i - 1}
  k.shift()
  k.each do |j|
    a[j] |=  1<<i
  end
end

a_pi = gets.split.map(&:to_i)
a_pi.each_with_index do |x, i|
  pi |= x << i
end

s = 0;
while(s < (1<<n)) do
  t = 0;
  for i in 0..(n-1) do
    if s[i] != 0
      t ^= a[i];
    end
  end
  ans += 1 if t == pi
  s += 1
end

puts ans