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::vec::Vec;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
    let (n, m) = {
      let v: Vec<usize> = read_vec();
      (v[0], v[1])
    };

    let mut ks: Vec<Vec<i64>> = Vec::new();
    for _ in 0..m {
      let v: Vec<i64> = read_vec();
      ks.push(v);
    }

    let p: Vec<i64> = read_vec();
    let ans = dfs(n as i64, Vec::new(), &ks, &p);
    println!("{}", ans);
}


fn dfs(n: i64, v: Vec<i64>, ks: &Vec<Vec<i64>>, p: &Vec<i64>) -> i64 {
  if n == 0 {
    return count(v, ks, p);
  }

  let mut v1 = v.clone();
  v1.push(0);
  let mut v2 = v.clone();
  v2.push(1);
  let c1 = dfs(n-1, v1, ks, p);
  let c2 = dfs(n-1, v2, ks, p);

  return c1 + c2;
}

fn count(v: Vec<i64>, ks: &Vec<Vec<i64>>, p: &Vec<i64>) -> i64 {
  let m: usize = ks.len();


  let mut count = 1;

  for i in 0..m {
    let mut temp = 0;
    for j in 1..ks[i][0] as usize + 1 {
      if v[ks[i][j] as usize - 1] == 0 {
        temp += 1;
      }
    }
    if temp % 2 != p[i] {
      count = 0;
    }
  }
  return count;
}