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 OCaml (4.02.3) (judge verdict: Accepted)

Scanf.scanf "%d %d" @@ fun n m ->
  let sss = Array.init m @@ fun _ ->
    let k = Scanf.scanf " %d" @@ (+) 0 in
    Array.init k @@ fun _ -> Scanf.scanf " %d" @@ fun s -> s - 1 in
  let ps = Array.init m @@ fun _ -> Scanf.scanf " %d" @@ (+) 0 in
  let array_count f = Array.fold_left (fun n x -> if f x then n + 1 else n) 0 in
  let array_for_all f xs = Array.length xs = array_count f xs in
  let flags = Array.make n false in
  let is_on i =
    let ss = sss.(i) in
    let c = ref 0 in
    for j = 0 to Array.length ss - 1 do
      if flags.(ss.(j)) then incr c
    done;
    !c mod 2 = ps.(i) in
    (* let c = array_count ((=) true) flags in c mod 2 = ps.(i) in *)
  let judge () = array_for_all is_on @@ Array.init m @@ (+) 0 in
  let ans = ref 0 in
  let rec find i =
    (* Array.iter (fun x -> print_string @@ if x then "true " else "false ") flags;
    print_newline (); *)
    (* Printf.printf "%d %d\n" i n; *)
    if i >= n then
      (if judge () then incr ans)
    else
      (flags.(i) <- true;
      find (i + 1);
      flags.(i) <- false;
      find (i + 1)) in
  find 0;
  Printf.printf "%d\n" !ans