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 (1.42.0) (judge verdict: Accepted)

#![allow(unused_imports)]

use proconio::{input, fastout};
use proconio::marker::*;
use nalgebra::max;


#[fastout]
fn main() {
    input! {
    n: usize
    }
    let mut an = vec![];
    let mut xyn = vec![];
    for _ in 0..n {
        input! {
        a: usize,
        xy: [(Usize1, usize); a]
        }
        an.push(a);
        xyn.push(xy);
    }
    let mut ans = 0;
    for i in 0..1 << n {
        let mut f = true;
        let honest = (i as usize).count_ones();
        for j in 0..n {
            let (a, xy) = (an[j], &xyn[j]);
            if i & 1 << j == 0 {continue}
            for k in 0..a {
                let (x, y) = xy[k];
                if (i >> x) % 2 != y {
                    f = false;
                    break;
                }
            }
            if !f { break; }
        }
        if !f { continue; }
        ans = max(honest, ans);
    }
    println!("{}", ans);
}