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::*;
use std::str::FromStr;

//https://qiita.com/tubo28/items/e6076e9040da57368845
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn rec(
    n: usize,
    m: usize,
    status: &mut Vec<usize>,
    mut ans: u32,
    switch: &Vec<Vec<usize>>,
    p: &Vec<u32>,
) -> u32 {
    //最深まで探索し、状態を判定する
    if status.len() == n {
        let mut light = vec![false; m];
        for i in 0..m {
            let mut on = 0;
            for j in switch[i].iter() {
                if (status[*j - 1]) == 1 {
                    on += 1;
                }
            }
            //点灯の確認
            //println!("on{}",on);
            if on % 2 == p[i]{
                light[i] = true;
            }
        }
        //全点灯の確認
        if light.iter().all(|c| *c == true) {
            ans += 1;
        }
        //println!("ans{} status{:?}",ans,status);
    } else {
        //再帰全探索
        for c in 0..2 {
            status.push(c);
            ans = rec(n, m, status, ans, switch, p);
            status.pop();
        }
    }

    return ans;
}
fn main() {
    let n: usize = read();
    let m: usize = read();
    let mut switch: Vec<Vec<usize>> = vec![vec![0; 0]; m]; //m個の空のvecを用意

    for i in 0..m {
        let k: usize = read();
        for _ in 0..k {
            let s: usize = read();
            switch[i].push(s);
        }
    }
    let p: Vec<u32> = (0..m).map(|_| read()).collect();

    let mut status = Vec::new();
    let ans = rec(n, m, &mut status, 0, &switch, &p);
    print!("{}", ans);
}