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)

let () =
    let split s =
        let rec loop i cur flag acc =
            if i = String.length s then (
                if flag then cur :: acc else acc
            ) else (
                match flag, s.[i] with
                | true,  ' ' -> loop (i + 1) 0 false (cur :: acc)
                | false, ' ' -> loop (i + 1) 0 false acc
                | true,  v   -> loop (i + 1) (cur * 10 + int_of_char v - 48) true acc
                | false, v   -> loop (i + 1) (cur * 10 + int_of_char v - 48) true acc
            )
        in
        List.rev (loop 0 0 false [])
    in
    let main () =
        let n, m = Scanf.sscanf (read_line ()) "%d %d" (fun n m -> n, m) in
        let arr = Array.init m (fun _ -> List.tl (split (read_line ()))) in
        let p = Array.of_list (split (read_line ())) in
        let sw = Array.make m 0 in
        for i = 0 to m - 1 do
            sw.(i) <- List.fold_left (fun acc v -> acc lor (1 lsl (v - 1))) 0 arr.(i)
        done;
        let count bits =
            let rec loop i acc =
                if i = 0 then acc else loop (i / 2) (acc + i land 1)
            in
            loop bits 0
        in
        let forall bits switch p =
            let rec loop i =
                if i < 0 then true else
                if count (switch.(i) land bits) mod 2 = p.(i) then loop (i - 1) else false
            in
            loop (m - 1)
        in
        let rec loop i acc =
            if i < 0 then acc else
                let acc = if forall i sw p then acc + 1 else acc in
                loop (i - 1) acc
        in
        let ans = loop ((1 lsl n) - 1) 0 in
        Printf.printf "%d\n" ans
    in
    main ()