Problem:
Score : 300 points

Problem Statement

There are N people numbered 1 to N. Each of them is either an honest person whose testimonies are always correct or an unkind person whose testimonies may be correct or not.

Person i gives A_i testimonies. The j-th testimony by Person i is represented by two integers x_{ij} and y_{ij}. If y_{ij} = 1, the testimony says Person x_{ij} is honest; if y_{ij} = 0, it says Person x_{ij} is unkind.

How many honest persons can be among those N people at most?

Constraints

All values in input are integers.

1 \leq N \leq 15

0 \leq A_i \leq N - 1

1 \leq x_{ij} \leq N

x_{ij} \neq i

x_{ij_1} \neq x_{ij_2} (j_1 \neq j_2)

y_{ij} = 0, 1

Input

Input is given from Standard Input in the following format:

N
A_1
x_{11} y_{11}
x_{12} y_{12}
:
x_{1A_1} y_{1A_1}
A_2
x_{21} y_{21}
x_{22} y_{22}
:
x_{2A_2} y_{2A_2}
:
A_N
x_{N1} y_{N1}
x_{N2} y_{N2}
:
x_{NA_N} y_{NA_N}

Output

Print the maximum possible number of honest persons among the N people.

Sample Input 1

3
1
2 1
1
1 1
1
2 0

Sample Output 1

2

If Person 1 and Person 2 are honest and Person 3 is unkind, we have two honest persons without inconsistencies, which is the maximum possible number of honest persons.

Sample Input 2

3
2
2 1
3 0
2
3 1
1 0
2
1 1
2 0

Sample Output 2

0

Assuming that one or more of them are honest immediately leads to a contradiction.

Sample Input 3

2
1
2 0
1
1 0

Sample Output 3

1

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

fn solve(xys: Vec<Vec<(usize, usize)>>) -> usize {
    let n = xys.len();
    'outer: for i in (0..2usize.pow(n as u32)).rev() {
        let mut aa: Vec<Option<usize>> = vec![None; n];
        let mut ans = 0;
        for j in 0..n {
            if (i & (1 << j)) > 0 {
                ans += 1;
                for &(x, y) in xys[j].iter() {
                    if let Some(yp) = aa[x - 1] {
                        if yp != y {
                            continue 'outer;
                        }
                    } else {
                        aa[x - 1] = Some(y);
                    }
                }
            } else {
                if let Some(yp) = aa[j] {
                    if yp != 0 {
                        continue 'outer;
                    }
                }
                aa[j] = Some(0);
            }
        }
        return ans;
    }

    0
}

#[test]
fn solve_test() {
    assert_eq!(solve(vec![vec![(2, 1)], vec![(1, 1)], vec![(2, 0)]]), 2);
}

#[allow(unused_macros)]
macro_rules! parse_line {
    ( $t:ty ) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            iter.next().unwrap().parse::<$t>().unwrap()
        }
    );

    ( $( $t:ty), +) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            ( $(iter.next().unwrap().parse::<$t>().unwrap()),* )
        }
    );
}

#[allow(unused_macros)]
macro_rules! read_line {
    () => {{
        let mut line = String::new();
        ::std::io::stdin().read_line(&mut line).unwrap();
        line.pop();
        line
    }};
}

#[allow(unused_macros)]
macro_rules! parse_vec {
    ( $t:ty ) => {{
        let mut line = String::new();
        ::std::io::stdin().read_line(&mut line).unwrap();
        let iter = line.split_whitespace();
        iter.map(|v| v.parse::<$t>().unwrap()).collect::<Vec<_>>()
    }};
}

fn main() {
    let n = parse_line!(usize);

    let mut xys = Vec::new();

    for _ in 0..n {
        let a = parse_line!(usize);

        let v = (0..a)
            .map(|_| parse_line!(usize, usize))
            .collect::<Vec<_>>();
        xys.push(v);
    }

    println!("{}", solve(xys));
}