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: Accepted)

use std::io::Read;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let mut z: Vec<Vec<(usize, usize)>> = vec![Vec::new(); n];
    for i in 0..n {
        let a: usize = itr.next().unwrap().parse().unwrap();
        for _ in 0..a {
            let x: usize = itr.next().unwrap().parse().unwrap();
            let y: usize = itr.next().unwrap().parse().unwrap();
            z[i].push((x - 1, y));
        }
    }

    let mut ans = 0;
    for bit in 0..1 << n {
        let mut ok: bool = true;
        'soto: for i in 0..n {
            if bit >> i & 1 == 1 {
                for &(x, y) in z[i].iter() {
                    if bit >> x & 1 != y {
                        ok = false;
                        break 'soto;
                    }
                }
            }
        }
        if ok {
            ans = std::cmp::max(ans, (bit as u32).count_ones());
        }
    }
    println!("{}", ans);
}