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::{stdin, Read, StdinLock};
use std::str::FromStr;

#[allow(dead_code)]
struct Scanner<'a> {
    cin: StdinLock<'a>,
}
#[allow(dead_code)]
impl<'a> Scanner<'a> {
    fn new(cin: StdinLock<'a>) -> Scanner<'a> {
        Scanner { cin: cin }
    }

    fn read<T: FromStr>(&mut self) -> Option<T> {
        let token = self.cin.by_ref().bytes().map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        token.parse::<T>().ok()
    }

    fn input<T: FromStr>(&mut self) -> T {
        self.read().unwrap()
    }

    fn vec<T: FromStr>(&mut self, len: usize) -> Vec<T> {
        (0..len).map(|_| self.input()).collect()
    }

    fn mat<T: FromStr>(&mut self, row: usize, col: usize) -> Vec<Vec<T>> {
        (0..row).map(|_| self.vec(col)).collect()
    }
}

fn main() {
    let cin = stdin();
    let cin = cin.lock();
    let mut sc = Scanner::new(cin);
    let n: i64 = sc.input();
    let m: i64 = sc.input();
    let mut arr = vec![Vec::<i64>::new(); m as usize];
    for i in 0..m {
        let k: i64 = sc.input();
        for _ in 0..k {
            arr[i as usize].push(sc.input());
        }
    }
    let ps: Vec<i64> = sc.vec(m as usize);

    let mut ans = 0;
    for mask in 0..(1 << n) {
        let mut ok = true;
        for i in 0..m {
            let mut cnt = 0;
            for sw in &arr[i as usize] {
                if mask >> (sw - 1) & 1 == 1 {
                    cnt += 1;
                }
            }
            if cnt % 2 != ps[i as usize] {
                ok = false;
            }
        }
        if ok {
            ans += 1;
        }
    }
    println!("{}", ans);
}