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::io::prelude::*;
use std::io::stdin;
use std::str::FromStr;

fn main() {
    let n: usize = read();
    let m: usize = read();
    let mut ss = vec![];

    for _ in 0..m {
        let k: usize = read();
        let s: Vec<usize> = (0..k).map(|_| read()).collect();
        ss.push(s);
    }
    let p: Vec<usize> = (0..m).map(|_| read()).collect();

    let mut switchs = vec![];
    let switch = vec![];
    make_switch_pattern(&mut switchs, &switch, n);

    let ans = switchs
        .iter()
        .filter(|switch| {
            ss.iter()
                .zip(p.iter())
                .filter(|&(s, p_i)| s.iter().filter(|s_i| !switch[**s_i - 1]).count() % 2 == *p_i)
                .count()
                == m
        })
        .count();

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

// true: on, false: off
fn make_switch_pattern(switchs: &mut Vec<Vec<bool>>, switch: &Vec<bool>, count: usize) {
    if count == 0 {
        switchs.push(switch.clone());
    } else {
        let mut switch1 = switch.clone();
        switch1.push(true);
        make_switch_pattern(switchs, &switch1, count - 1);
        let mut switch2 = switch.clone();
        switch2.push(false);
        make_switch_pattern(switchs, &switch2, count - 1);
    }
}

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|b| b.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}