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 Lua (5.3.2) (judge verdict: Accepted)

local Nsw, Nli = io.read("n", "n")
local switches = {}
local onconds = {}
for i=1,Nli do
    local k = io.read("n")
    local swi = {}
    switches[i] = swi
    for j=1,k do
        swi[j] = io.read("n")
    end
end
for i=1,Nli do
    onconds[i] = io.read("n")
end
-- Cartesian product
local function carprod(seqs)
    return coroutine.wrap(function()
        if #seqs == 0 then
            coroutine.yield({})
            return
        end
        local s1 = seqs[1]
        for _,v in ipairs(s1) do
            for t in carprod({select(2, table.unpack(seqs))}) do
                coroutine.yield({v, table.unpack(t)})
            end
        end
    end)
end

local zo = {0,1}
local set = {}
for i=1,Nsw do
    set[i] = zo
end
local ans = 0
for ss in carprod(set) do -- 1024 times
    --print(table.concat(ss, ""))
    local all_on = true
    for i=1,Nli do -- 10 times
        local swi = switches[i]
        local sum = 0
        for k,v in ipairs(swi) do -- 10 times
            if ss[v] == 1 then
                sum = sum + 1
            end
        end
        if sum % 2 == onconds[i] then
        else
            all_on = false
            break
        end
    end
    if all_on then
        ans = ans + 1
    end
end
print(ans)