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 Common Lisp (SBCL 2.0.3) (judge verdict: Accepted)

(let* ((n (read))
       (m (read))
       (k (make-array (list (1+ m))))
       (s (make-array (list (1+ m) (1+ n))))
       (p (make-array (list (1+ m))))
       (ans 0))
  (loop :for i :from 1 :to m
        :do (setf (aref k i) (read))
        :do (loop :for j :from 1 :to (aref k i)
                  :do (setf (aref s i j) (read))))
  (loop :for i :from 1 :to m
        :do (setf (aref p i) (read)))
  (defun check (cnt sw)
    (if (= cnt (+ n 1))
        (if (= (loop :for i :from 1 :to m
                     :if (= (mod (loop :for j :from 1 :to (aref k i)
                                       :sum (aref sw (aref s i j))) 
                                 2)
                            (aref p i))
                     :sum 1)
               m)
              (incf ans))
        (loop :for a :from 0 :to 1
              :do (setf (aref sw cnt) a)
              :do (check (1+ cnt) sw))))
  (check 1 (make-array (list (1+ n))))
  (format t "~A~%" ans))