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)

#[cfg(debug_assertions)]
include! {"./debug.rs"}

#[cfg(not(debug_assertions))]
macro_rules! debug {
    ($($arg:expr),*) => {};
}

macro_rules! read {
    ([$t:ident] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ([$t:ident]) =>
        (rl().split_whitespace().map(|w| read_type!(w, $t)).collect::<Vec<_>>());
    (_, [$t:ident] ; $n:expr) =>
        ((0..$n).map(|_| read!(_, [$t])).collect::<Vec<_>>());
    (_, [$t:ident]) =>
        (rl().split_whitespace().skip(1).map(|w| read_type!(w, $t)).collect::<Vec<_>>());
    ($($t:ident),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ($t:ident) => (read_type!(rl(), $t));
    ($($t:ident),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(read_type!(w.next().unwrap(), $t)),*)
    }};
}

macro_rules! read_type {
    ($s:expr, chars) => {
        $s.chars().collect::<Vec<char>>()
    };
    ($s:expr, usize1) => {
        read_type!($s, usize) - 1
    };
    ($s:expr, $t:ty) => {
        $s.parse::<$t>().unwrap()
    };
}

fn rl() -> String {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    buf.trim_right().to_owned()
}

////////////////////////////////////////////////////////////////////////////////

fn main() {
    let (n, m) = read!(usize, usize);
    let s = read!(_, [usize1]; m);
    let p = read!([usize]);

    let s = s
        .iter()
        .map(|v| v.iter().map(|x| 1usize << x).fold(0usize, |x, y| x | y))
        .collect::<Vec<_>>();

    let ans = (0..1 << n)
        .filter(|sw| {
            (0..m).all(|lgt| {
                let l = sw & s[lgt];
                (0..n).filter(|i| l & (1usize << i) != 0).count() % 2 == p[lgt]
            })
        })
        .count();

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