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::collections::HashMap;
use std::io::*;
use std::str::FromStr;

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 main() {
    // switch num
    let switch_num: i32 = read();
    // bulb num
    let bulb_num: i32 = read();
    let mut map = HashMap::new();
    // bi: bulb index
    for bi in 1..bulb_num + 1 {
        let k: i32 = read();
        for _ in 1..k + 1 {
            // si: switch index
            let si: i32 = read();
            let entry = map.entry(si).or_insert(vec![]);
            entry.push(bi)
        }
    }
    let mut vec = vec![true];
    for _ in 1..bulb_num + 1 {
        vec.push(read::<i32>() == 0);
    }
    println!("{}", dfs(1, switch_num, &map, vec, 0));
}

fn dfs(
    depth: i32,
    last: i32,
    map: &HashMap<i32, Vec<i32>>,
    switch_state: Vec<bool>,
    ans: i32,
) -> i32 {
    if depth == last + 1 {
        if switch_state.iter().all(|&x| x == true) {
            return ans + 1;
        }
        return ans;
    }
    let mut switch_state_flip = switch_state.clone();
    match map.get(&depth) {
        Some(vec) => {
            for &i in vec {
                switch_state_flip[i as usize] ^= true;
            }
        }
        None => {}
    };
    let ans1 = dfs(depth + 1, last, &map, switch_state_flip, ans);
    let ans2 = dfs(depth + 1, last, &map, switch_state, ans);
    ans1 + ans2
}