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)

#![allow(bare_trait_objects)]
#![allow(unused_imports)]
#![allow(dead_code)]

use std::collections::{HashMap, HashSet};
use std::fmt::Debug;
use std::io::BufRead;
use std::io::Stdin;
use std::io::StdinLock;
use std::str::FromStr;

pub struct Scanner<'a> {
    // stdin: Stdin,
    buf_read: Box<BufRead + 'a>,
    tokens: Vec<String>,
}

impl<'a> Scanner<'a> {
    pub fn new(buf_read: StdinLock<'a>) -> Self {
        Scanner {
            buf_read: Box::new(buf_read),
            tokens: Vec::new(),
        }
    }

    // pub fn from_stdin() -> Self {
    //     let stdin = Box::leak(Box::new(std::io::stdin()));
    //     Self {
    //         buf_read: Box::new(stdin.lock()),
    //         tokens: Vec::new(),
    //     }
    // }

    pub fn next(&mut self) -> String {
        while self.tokens.is_empty() {
            let mut s = String::new();
            if self.buf_read.read_line(&mut s).is_err() {
                panic!()
            }
            self.tokens = s
                .split_whitespace()
                .filter(|s| !s.is_empty())
                .rev()
                .map(str::to_string)
                .collect();
        }
        self.tokens.pop().unwrap()
    }

    pub fn next_g<T: FromStr>(&mut self) -> T
    where
        T::Err: Debug,
    {
        self.next().parse().unwrap()
    }

    pub fn usize(&mut self) -> usize {
        self.next_g()
    }

    pub fn i64(&mut self) -> i64 {
        self.next_g()
    }
}

fn main() {
    let stdin = std::io::stdin();
    let mut sc = Scanner::new(stdin.lock());

    let n = sc.usize();
    let m = sc.usize();

    let mut s = vec![0; n];
    for i in 0..m {
        let k = sc.usize();
        for _ in 0..k {
            s[sc.usize() - 1] |= 1 << i;
        }
    }

    let p = (0..m).map(|_| sc.usize()).collect::<Vec<_>>();

    let mut ans = 0;
    for i in 0..1 << n {
        let mut q = 0;
        for j in 0..n {
            if (i >> j & 1) == 1 {
                q ^= s[j];
            }
        }
        if (0..m).all(|b| (q >> b & 1) == p[b]) {
            ans += 1;
        }
    }
    println!("{}", ans);
}