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)

n, m = gets.split.map(&:to_i)
patterns = []

def dfs(array, patterns, limit)
  if array.length == limit
    patterns << array
  else
    dfs(array + [0], patterns, limit)
    dfs(array + [1], patterns, limit)
  end
  if array.empty?
    patterns
  end
end
dfs([], patterns, n)
# p patterns # => [[0, 0], [0, 1], [1, 0], [1, 1]]

rules = []
m.times do
  k, *s = gets.split.map(&:to_i)
  rules << s
end
ps = gets.split.map(&:to_i)
ps.each_with_index do |pnum, index|
  judge = (pnum == 0 ? :even : :odd)
  rules[index] = {switches: rules[index], judge: judge}
end
# p rules # => [{:switches=>[1, 2], :judge=>:even}, {:switches=>[2], :judge=>:odd}]

rules.each do |rule|
  break if patterns.empty?
  patterns.map! {|pattern|
    count = 0
    rule[:switches].each do |switch|
      count += pattern[switch-1]
    end
    if rule[:judge] == :even
      count.even? ? pattern : nil
    elsif rule[:judge] == :odd
      count.odd?  ? pattern : nil
    end
  }.compact!
end
# p patterns
puts patterns.count