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 Rust (1.15.1) (judge verdict: Accepted)

use std::str::FromStr;
use std::fmt::Debug;
use std::cmp;

#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

#[allow(dead_code)]
fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn main(){
    let line: Vec<usize> = read_as_vec();
    let (n, m) = (line[0], line[1]);

    let mut s: Vec<usize> = vec![0;m];
    for i in 0..m{
        let line: Vec<usize> = read_as_vec();
        let switches = line.into_iter().skip(1).collect::<Vec<usize>>();
        for switch in switches{
            s[i] |= 1 << (switch-1);
        }
    }

    let p: Vec<usize> = read_as_vec();
    let mut counter = 0;
        
    for cur_switch in 0..(2<<(n-1)){
        let mut is_on: Vec<bool> = vec![false;m];
        //println!("current switch status: {:b}", cur_switch);
        for i in 0..m{
            let filtered_switch = cur_switch & s[i];
            //println!("filtered switch: {:b}", filtered_switch);
            //let num_of_on = filtered_switch.to_string().chars().filter(|x| *x == '1').count();
            let num_of_on = format!("{:b}", filtered_switch).chars().filter(|x| *x == '1').count();
            //println!("num_of_on = {}", num_of_on);
            if num_of_on%2 == p[i]{
                is_on[i] = true;
            }
        }

        //println!("result; {:?}", is_on);
        if is_on.into_iter().all(|x| x){
            counter += 1;
        }
    }

    println!("{}", counter);
}