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::*;

fn count(comb: usize, s_arr: &Vec<usize>) -> usize {
    let mut result = 0;
    for s in s_arr {
        if (comb & (1usize << s)) != 0 {
            result += 1;
        }
    }
    result
}

fn solve(n: usize, m: usize, s_arr_arr: Vec<Vec<usize>>, p: Vec<usize>) ->
 usize {
    let mut comb: usize = 0;
    let end: usize = 1 << n;
    let mut result = 0;
    while comb < end {
        if (0..m).map(|i| {
            count(comb, &s_arr_arr[i]) % 2 == p[i]
        }).all(|a| { a })
        {
            result += 1;
        }
        comb += 1;
    }
    result
}

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();

    let n: usize = iter.next().unwrap().parse().unwrap();
    let m: usize = iter.next().unwrap().parse().unwrap();
    let s: Vec<Vec<usize>> = (0..m).map(|_| {
        let k: usize = iter.next().unwrap().parse().unwrap();
        (0..k).map(|_| {
            iter.next().unwrap().parse::<usize>().unwrap() - 1usize
        }).collect()
    }).collect();
    let p: Vec<usize> = (0..m).map(|_| {
        iter.next().unwrap().parse().unwrap()
    }).collect();
    println!("{}", solve(n, m, s, p));
}