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 JavaScript (node.js v5.12) (judge verdict: Accepted)

'use strict';(function(input){
  function count(pi,A,B){
    var C=A&B;
    var counter=pi;
    while(C>0){
      counter+=(C%2);
      C=(C>>1);
    }
    return ((counter%2)===0);
  }
  
  input=input.trim().split(/\n+/);
  input=input.map(str=>str.split(/\s+/).map(x=>x-0));
  const N=input[0][0];
  const M=input[0][1];
  const p=input.pop();
  for(let i=1;i<=M;i++)
    input[i].shift();
  const arr=new Array(1<<N).fill('filler').map((x,i)=>i);
  const mask=[];
  for(let i=1;i<=M;i++)
    mask.push(input[i].reduce((acc,cur)=>acc+(1<<(cur-1)),0));
  mask.forEach(function(A,i){
    let j=0;
    while(j<arr.length){
      if(!count(p[i],A,arr[j])){
        arr.splice(j,1);
      }else{
        j++;
      }
    }
  });
  console.log(arr.length);
})(require('fs').readFileSync('/dev/stdin','utf8'));